In a metric space you never need more than a list. Around any point, the balls of radius $1$, $\frac12$, $\frac13$, and so on do every job a neighborhood could do, and sequences find every point of a closure. This page asks which spaces get away with "countably many" too, and what it buys them.
Then the other half of the title. Hausdorff spaces separate two points with open sets. We'll ask for more: separate a point from a closed set, then two closed sets from each other. The two halves meet at the end, where a countable basis turns the weaker kind of separation into the stronger one. Practice sits right after each idea, and the Mixed Practice at the bottom takes that hint away.
Start small, at one point. A neighborhood of $x$ can be as weird as you like, but often a short list of standard ones is enough: every neighborhood contains one of them. When that list can be countable, we say so.
Let $X$ be a topological space and $x \in X$. $X$ is said to have a countable basis at $x$ if there exists a countable collection $\mathcal{B}_x$ of neighborhoods of $x$ such that each neighborhood of $x$ contains at least one of the elements of $\mathcal{B}_x$.
One point isn't the goal, though. We want the list at every point.
$X$ is said to be first-countable if it has a countable basis at each of its points.
The basic example is every metric space, and the second one is the lower limit line, where the list has to lean right.
Show that every metrizable space is first-countable, and that $\mathbb{R}_l$ is first-countable.
Let $d$ be a metric for $X$ and $x \in X$. Take $\mathcal{B}_x = \{B_d(x, \frac1n) : n \in \mathbb{N}\}$. Each ball is a neighborhood of $x$. If $U$ is any neighborhood of $x$, it contains some ball $B_d(x, \varepsilon)$, and choosing $n$ with $\frac1n < \varepsilon$ gives $B_d(x, \frac1n) \subseteq U$.
In $\mathbb{R}_l$, take $\mathcal{B}_x = \{[x, x + \frac1n) : n \in \mathbb{N}\}$. These are basis elements, so they're neighborhoods of $x$. A neighborhood $U$ of $x$ contains a basis element $[a, b)$ with $a \le x < b$, so it contains $[x, b)$, and that contains $[x, x + \frac1n)$ once $\frac1n < b - x$.
Here's the metric case as a picture. Whatever blob of a neighborhood you draw around $x$, the shrinking balls eventually fit inside it.
The list has to work for every neighborhood, including the skinny ones. Try it yourself. Drag and squeeze the neighborhood, and the readout finds the first set on the list that fits. In $\mathbb{R}_l$, watch which list survives.
So why care? Because a countable list at $x$ is exactly what lets sequences do the work they do in metric spaces.
On Metric Topology, the Sequence Lemma said: if a sequence in $A$ converges to $x$, then $x \in \overline{A}$, and the converse holds when $X$ is metrizable. The theorem below trades "metrizable" for "first-countable", which is all the proof ever used.
Let $X$ be a topological space.
Proof. (1) If $x_n \to x$ with every $x_n \in A$, then every neighborhood of $x$ contains some $x_n$, a point of $A$, so $x \in \overline{A}$.
Conversely, let $X$ be first-countable and $x \in \overline{A}$. Let $\mathcal{B}_x = \{B_1, B_2, \ldots\}$ be a countable basis at $x$, and put $C_n = B_1 \cap \cdots \cap B_n$. Each $C_n$ is a neighborhood of $x$, and $x \in \overline{A}$, so we can pick $x_n \in C_n \cap A$. Given a neighborhood $U$ of $x$, some $B_N \subseteq U$, and for $n \ge N$ we get $x_n \in C_n \subseteq B_N \subseteq U$. So $x_n \to x$.
(2) Let $f$ be continuous, $x_n \to x$, and $V$ a neighborhood of $f(x)$. Then $f^{-1}(V)$ is a neighborhood of $x$, so $x_n \in f^{-1}(V)$, that is $f(x_n) \in V$, for all $n$ from some point on.
Conversely, let $X$ be first-countable and suppose $f(x_n) \to f(x)$ whenever $x_n \to x$. Let $V$ be closed in $Y$; we show $f^{-1}(V)$ is closed. Let $x \in \overline{f^{-1}(V)}$. By (1), some sequence $x_n$ in $f^{-1}(V)$ converges to $x$. Then $f(x_n) \to f(x)$ with every $f(x_n) \in V$, so by the first half of (1), $f(x) \in \overline{V} = V$. So $x \in f^{-1}(V)$, and $f^{-1}(V)$ is closed. Preimages of closed sets are closed, so $f$ is continuous. $\blacksquare$
The trick in the proof is the nesting: $C_n$ shrinks, so once a term is inside $U$, every later term is too. Here it is, one $C_n$ at a time.
Without first-countability, sequences really can miss a closure point. The cleanest example is a topology where countable sets are tiny.
Give $\mathbb{R}$ the countable complement topology: the open sets are $\emptyset$ and the sets whose complement is countable. Show that $0 \in \overline{\mathbb{R} \setminus \{0\}}$, but no sequence in $\mathbb{R} \setminus \{0\}$ converges to $0$.
A neighborhood of $0$ has a countable complement, so it contains uncountably many points, and some of them aren't $0$. So every neighborhood of $0$ meets $\mathbb{R} \setminus \{0\}$, and $0$ is in the closure.
Now let $x_n$ be any sequence with every $x_n \neq 0$. The set $C = \{x_1, x_2, \ldots\}$ is countable, so $U = \mathbb{R} \setminus C$ is open, and it contains $0$. No term of the sequence is in $U$, so $x_n \not\to 0$. By the theorem, this space can't be first-countable.
You can see the failure directly. Somebody hands you a countable list of neighborhoods of $x$, and you have to find a neighborhood that contains none of them. In some spaces you always can. Try it.
In the finite complement topology the winning move is the same as in the countable complement one: a countable list only ever mentions countably many points, so pick one it never mentions and throw it out. Your turn: pick the list that really is a countable basis at the point.
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First-countable means a countable list at each point. The stronger version asks for one countable list that works everywhere at once: a countable basis for the whole topology.
If the topological space $X$ has a countable basis for its topology, then $X$ is said to be second-countable.
A second-countable space is first-countable: the basis elements that contain $x$ form a countable basis at $x$. And the first example is the one you'd guess.
Show that $\mathbb{R}$ and $\mathbb{R}^n$ are second-countable.
In $\mathbb{R}$, take all open intervals $(a, b)$ with rational endpoints. There are countably many, since $\mathbb{Q} \times \mathbb{Q}$ is countable. They form a basis: if $U$ is open and $x \in U$, some $(c, d)$ has $x \in (c, d) \subseteq U$, and rationals $a, b$ with $c < a < x < b < d$ give $x \in (a, b) \subseteq U$.
In $\mathbb{R}^n$, take all products $(a_1, b_1) \times \cdots \times (a_n, b_n)$ of open intervals with rational endpoints. That's countably many, and the same squeeze, done in each coordinate, shows they form a basis.
Both properties pass down to subspaces and across countable products, and the proof is just "intersect with $A$" and "use finitely many coordinates".
A subspace of a first-countable space is first-countable, and a countable product of first-countable spaces is first-countable. A subspace of a second-countable space is second-countable, and a countable product of second-countable spaces is second-countable.
Proof. Subspaces: if $\mathcal{B}$ is a countable basis for $X$ and $A \subseteq X$, then $\{B \cap A : B \in \mathcal{B}\}$ is a countable basis for $A$, because the open sets of $A$ are the sets $U \cap A$ with $U$ open in $X$. The same sentence with $\mathcal{B}_x$ at a point $x \in A$ handles first-countability.
Products: let $\mathcal{B}_i$ be a countable basis for $X_i$, $i \in \mathbb{N}$. For each $N$, consider the products $U_1 \times \cdots \times U_N \times X_{N+1} \times X_{N+2} \times \cdots$ with each $U_i \in \mathcal{B}_i \cup \{X_i\}$. There are countably many for each $N$, so countably many in all. A basis element of the product topology around a point $x$ has the form $V_1 \times \cdots \times V_N \times X_{N+1} \times \cdots$ with each $V_i$ open; choosing $U_i \in \mathcal{B}_i$ with $x_i \in U_i \subseteq V_i$ for $i \le N$ gives one of our sets containing $x$ and lying inside it. So they form a countable basis. At a single point $x$, the same construction with countable bases at each $x_i$ gives a countable basis at $x$. $\blacksquare$
The product half is worth trying once by hand, on the most important countable product there is.
$\mathbb{R}^\omega$, with the product topology, has a countable basis. Find one.
Take all products $U_1 \times U_2 \times \cdots$ where finitely many $U_n$ are open intervals with rational endpoints and all the others are $\mathbb{R}$. For each finite set of coordinates there are countably many choices, and there are countably many finite sets of coordinates, so the collection is countable.
It's a basis: a product basis element around $x$ is $V_1 \times \cdots \times V_N \times \mathbb{R} \times \cdots$ with each $V_n$ open, and for $n \le N$ there is a rational interval $U_n$ with $x_n \in U_n \subseteq V_n$. Then $U_1 \times \cdots \times U_N \times \mathbb{R} \times \cdots$ is in the collection, contains $x$, and lies inside the basis element. $\blacksquare$
Now the spaces that are first-countable and not second-countable. The lower limit line is the first. Every point needs its own private basis element.
Show that $\mathbb{R}_l$ is not second-countable.
Let $\mathcal{B}$ be any basis for $\mathbb{R}_l$. For each $x$, the set $[x, x + 1)$ is open, so some $B_x \in \mathcal{B}$ has $x \in B_x \subseteq [x, x + 1)$. Then $x$ is the smallest element of $B_x$. So if $x \neq y$, then $B_x \neq B_y$: they have different smallest elements. That gives one basis element for every real number, so $\mathcal{B}$ is uncountable.
Watch the points claim their sets. Each $x$ needs a set that starts exactly at $x$, and no two points can share one.
First-countable, second-countable... I'm waiting for third-countable. It doesn't exist. The franchise ended at two.
The second example lives in a metric space, so it's first-countable for free. The trouble is that it's too spread out.
(A space that is first-countable but not second-countable.) Consider $\mathbb{R}^\omega$ with the uniform topology, given by the uniform metric $\overline{\rho}$. Show that it is first-countable and not second-countable.
This topology is metrizable, so $\mathbb{R}^\omega$ with the uniform topology is first-countable. However, this space is not second-countable. To see this, consider the set $E \subseteq \mathbb{R}^\omega$ consisting of all sequences of $0$s and $1$s. This set is uncountable, and the subspace topology on it is the discrete topology, since $\overline{\rho}\big((x_n)_{n=1}^\infty, (y_n)_{n=1}^\infty\big) = 1$ for all pairs of distinct sequences in $E$. $\mathbb{R}^\omega$ with the uniform topology therefore cannot have a countable basis for its topology.
Each $0$–$1$ sequence sits alone in its own ball of radius $\frac12$, and there are uncountably many of them. Here they are as strips of bars, pulling apart.
The last sentence of that example skipped a step. Fill it in.
Explain why the final assertion is true: why can't $\mathbb{R}^\omega$ with the uniform topology have a countable basis?
If it had a countable basis, then by the theorem above so would its subspace $E$. But $E$ is discrete, so each one-point set $\{e\}$ is open. A basis element containing $e$ and lying inside $\{e\}$ must be $\{e\}$ itself, so every basis for $E$ contains every one-point set $\{e\}$. There are uncountably many $e \in E$ (a $0$–$1$ sequence is a binary expansion, and a diagonal argument shows they can't be listed), so no basis for $E$ is countable. $\blacksquare$
Second-countability has two famous consequences, and each one gets a name of its own. First, a set that comes close to everything.
A subset $A$ of a topological space $X$ is said to be dense in $X$ if $\overline{A} = X$.
So $A$ is dense when every nonempty open set meets it. Here's the model to keep in mind, and the two ways a set can fail.
Decide which of these are dense in $\mathbb{R}$: $\mathbb{Q}$; $\mathbb{R} \setminus \mathbb{Q}$; $\mathbb{Z}$; $\mathbb{Q} \cap [0, 1]$.
Every nonempty open set contains an interval $(a, b)$, so it's enough to check intervals.
The first two are dense, but only $\mathbb{Q}$ is countable, and that's the property worth a name.
A topological space $X$ is separable if it has a countable dense subset.
The second consequence is a weak cousin of compactness. Compact asks every open cover for a finite subcover; this one settles for a countable one.
A topological space $X$ is Lindelöf if every open cover of $X$ contains a countable subcollection that covers $X$.
Two quick sources of Lindelöf spaces, and one space that isn't.
Show that every compact space and every countable space is Lindelöf, and that $\mathbb{R}$ with the discrete topology is not.
A compact space has a finite subcover of every open cover, and finite is countable. In a countable space, choose one member of the cover containing each point: countably many choices.
In $\mathbb{R}$ with the discrete topology, the one-point sets $\{x\}$ are open and cover the space. Each point lies in exactly one of them, so any subcover has to keep all uncountably many.
Now the theorem that ties them to a countable basis. Both proofs are "one choice per basis element".
Suppose that $X$ is a second-countable topological space. The following hold.
Proof. Let $\{B_n\}$ be a countable basis.
(1) Let $\mathcal{A}$ be an open cover. Call $n$ good if $B_n$ lies inside some member of $\mathcal{A}$, and for each good $n$ choose one such member $A_n$. The collection $\{A_n : n \text{ good}\}$ is countable, since it has at most one member for each $n$. It covers $X$: given $x \in X$, pick $A \in \mathcal{A}$ containing $x$. Since $A$ is open, some $B_n$ has $x \in B_n \subseteq A$, so $n$ is good, and $x \in B_n \subseteq A_n$.
(2) Select a point $x_n$ from each nonempty basis element $B_n$. Let $D$ be the set consisting of the points $x_n$. Then $D$ is dense in $X$. To see this, let $x \in X$. Given any neighborhood $V$ of $x$, $V$ contains some basis element $B_n$ that contains $x$, and $B_n$ contains a point of $D$. So $x \in \overline{D}$. $\blacksquare$
Part (2) as a picture: one dot per basis element, and then every open set, wherever you put it, catches a dot.
Part (1) as a picture: the cover can have uncountably many sets, but each one only gets kept if it's the chosen home of some basis element, and there are only countably many of those.
Neither consequence goes back the other way, and the lower limit line shows it twice. It's separable, because every $[a, b)$ contains a rational, so $\mathbb{Q}$ is dense. It's not second-countable, as we saw. And it's Lindelöf, which takes a real argument.
Show that $\mathbb{R}_l$ is Lindelöf.
Proof. It's enough to handle covers by basis elements: given any open cover, write each member as a union of basis elements; a countable subcover of that refined cover gives a countable subcover of the original, by choosing for each kept basis element one member that contains it.
So let $\mathcal{A} = \{[a_\alpha, b_\alpha)\}$ cover $\mathbb{R}_l$. Let $C = \bigcup_\alpha (a_\alpha, b_\alpha)$, which is open in the usual topology of $\mathbb{R}$. As a subspace of $\mathbb{R}$, $C$ is second-countable, so the cover of $C$ by the intervals $(a_\alpha, b_\alpha)$ has a countable subcover.
It remains to cover $\mathbb{R} \setminus C$. Each $x \notin C$ lies in some $[a_\alpha, b_\alpha)$ but in no $(a_\alpha, b_\alpha)$, so $x = a_\alpha$ for some $\alpha$; choose such an $\alpha$ and a rational $q_x \in (x, b_\alpha)$. If $x < y$ are both outside $C$, then $y \notin (x, b_\alpha) \subseteq C$, so $q_x < y < q_y$. So $x \mapsto q_x$ is one-to-one, and $\mathbb{R} \setminus C$ is countable. One member of $\mathcal{A}$ for each of its points, together with the countable subcover of $C$, is a countable subcover. $\blacksquare$
In a metric space, though, the three ideas collapse into one. That's worth knowing, because it's how you usually prove a metric space is second-countable.
Show that every separable metric space is second-countable.
Proof. Let $D$ be a countable dense subset of the metric space $(X, d)$. The balls $B_d(q, \frac1m)$ with $q \in D$ and $m \in \mathbb{N}$ form a countable collection. Let $U$ be open and $x \in U$, with $B_d(x, \varepsilon) \subseteq U$. Choose $m$ with $\frac2m < \varepsilon$, and, since $D$ is dense, a point $q \in D \cap B_d(x, \frac1m)$. Then $x \in B_d(q, \frac1m)$, and by the triangle inequality $B_d(q, \frac1m) \subseteq B_d(x, \frac2m) \subseteq U$. So these balls form a countable basis. $\blacksquare$
Two questions to check the new words. First, which set is dense? Watch the topology: the same set can be dense in one and not the other.
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Then all four countability properties at once. Each question names a combination; find the space that has it.
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One can strengthen the Hausdorff separation axiom in many ways. Before we strengthen it, here's the rung just below it, the one that only asks points to be closed.
Hausdorff Spaces mentioned the whole ladder $T_0, T_1, T_2, \ldots$ and proved that one-point sets are closed in a Hausdorff space. This section climbs the ladder: $T_1$ below Hausdorff, then regular and normal above it.
A topological space $X$ is a $T_1$ space if every one-point set $\{x\}$ is closed in $X$.
There's an equivalent form that sounds more like Hausdorff, with the "disjoint" taken out.
Show that $X$ is $T_1$ if and only if for each pair of distinct points $x, y$, each one has a neighborhood that doesn't contain the other.
Proof. ($\implies$) If $\{y\}$ is closed, $X \setminus \{y\}$ is a neighborhood of $x$ that doesn't contain $y$; swap the roles for the other one.
($\impliedby$) Fix $x$. Each $y \neq x$ has a neighborhood $V_y$ that doesn't contain $x$. Then $X \setminus \{x\} = \bigcup_{y \neq x} V_y$ is open, so $\{x\}$ is closed. $\blacksquare$
Every Hausdorff space is $T_1$. The converse fails, and the counterexample is a space where open sets are too big to ever be disjoint.
Show that $\mathbb{R}$ with the finite complement topology is $T_1$ but not Hausdorff.
The closed sets are $\mathbb{R}$ and the finite sets, so every one-point set is closed: $T_1$.
Let $U = \mathbb{R} \setminus F$ and $V = \mathbb{R} \setminus G$ be nonempty open sets, with $F$, $G$ finite. Then $U \cap V = \mathbb{R} \setminus (F \cup G)$, which is nonempty because $F \cup G$ is finite and $\mathbb{R}$ is not. So no two points have disjoint neighborhoods: not Hausdorff.
Now up the ladder. Hausdorff separates a point from a point. The next two rungs replace one point, then both, by a closed set. Following Munkres, both definitions also ask for one-point sets to be closed.
Suppose that one-point sets are closed in $X$. Then $X$ is said to be regular if for each pair consisting of a point $x$ and a closed set $B$ disjoint from $\{x\}$, there exist disjoint open sets containing $x$ and $B$, respectively.
And the strongest rung on this page, both sides closed sets.
Suppose that one-point sets are closed in $X$. Then $X$ is said to be normal if whenever $A$ and $B$ are closed sets in $X$ with $A \cap B = \emptyset$, there exist disjoint open sets $U$ and $V$ with $A \subseteq U$ and $B \subseteq V$.
That "one-point sets are closed" clause looks like fine print, but it's what makes the rungs line up, and not every book includes it.
Watch the convention. Some books, including the course notes this page comes from, leave "one-point sets are closed" out of "normal" (and out of "regular"), and call a normal space with closed points a T4 space (a regular one T3). Without closed points, normal doesn't even imply Hausdorff: the two-point space with open sets ∅, {1}, {0, 1} has no two disjoint nonempty closed sets, so it's "normal" with nothing to check, and it isn't Hausdorff. This site follows Munkres, as on the Stone–Čech page.
With closed points built in, the rungs really are a ladder: normal implies regular (a point is a closed set), regular implies Hausdorff (so is the other point), and Hausdorff implies $T_1$. Here's what each rung separates.
There's a second way to say regular and normal, with one open set instead of two. It's the form that proofs actually use, and the Urysohn lemma will lean on it hard.
Let $X$ be a topological space in which one-point sets are closed.
Proof. (2) ($\implies$) The set $B = X \setminus U$ is closed and disjoint from $A$. By normality there are disjoint open sets $V \supseteq A$ and $W \supseteq B$. Then $V \subseteq X \setminus W$, a closed set, so $\overline{V} \subseteq X \setminus W \subseteq X \setminus B = U$.
($\impliedby$) Let $A$, $B$ be disjoint closed sets, and put $U = X \setminus B$, an open set containing $A$. Choose an open $V \supseteq A$ with $\overline{V} \subseteq U$. Then $V$ and $X \setminus \overline{V}$ are disjoint open sets containing $A$ and $B$.
(1) The same argument, with the closed set $A$ replaced by the point $x$. $\blacksquare$
So regular means you can always shrink a neighborhood a little and keep its closure inside. That's exactly what fails in the classic Hausdorff space that isn't regular.
Recall $\mathbb{R}_K$: $\mathbb{R}$ with the basis of open intervals $(a, b)$ and sets $(a, b) \setminus K$, where $K = \{\frac1n : n \in \mathbb{N}\}$. Show that $\mathbb{R}_K$ is Hausdorff but not regular.
$\mathbb{R}_K$ is finer than $\mathbb{R}$, so disjoint intervals still separate points: Hausdorff, and so one-point sets are closed.
$K$ is closed: its complement is the union of the basis elements $(-n, n) \setminus K$. And $0 \notin K$. Suppose $U \ni 0$ and $V \supseteq K$ were disjoint open sets. $U$ contains a basis element around $0$. It can't be an interval $(a, b)$, which would contain points $\frac1n$ of $K \subseteq V$; so it's $(a, b) \setminus K$. Choose $n$ with $\frac1n \in (a, b)$. The open set $V$ contains a basis element around $\frac1n$, and that must be an interval $(c, d)$, since $(c, d) \setminus K$ leaves $\frac1n$ out. Now $(c, d) \cap (a, b)$ is an interval around $\frac1n$, and it contains points not in $K$. Such a point is in both $U$ and $V$. Contradiction.
Here's the failure as a picture. Every open set around $0$ is an interval with the points of $K$ plucked out, and its closure puts them right back.
Now you've seen every rung fail once. Sort the whole zoo onto the ladder, then try the question.
Each question names a rung a space reaches and one it misses. Find the space.
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One property that does pass down easily: closed subspaces of normal spaces stay normal.
Show that a closed subspace of a normal space is normal.
Proof. Let $Y$ be closed in the normal space $X$. One-point sets are closed in $X$, so in $Y$. Let $A$, $B$ be disjoint closed sets of $Y$. Since $Y$ is closed in $X$, they are closed in $X$, so there are disjoint open sets $U \supseteq A$, $V \supseteq B$ of $X$. Then $U \cap Y$ and $V \cap Y$ are disjoint open sets of $Y$ containing $A$ and $B$. $\blacksquare$
Normal is the strongest rung, so it's the hardest to check. Three big families qualify, and the first is every space you've ever met with a distance in it.
Every metrizable space is normal.
Proof. Let $X$ be metrizable with metric $d$. One-point sets are closed in a metric space. Let $A$ and $B$ be disjoint closed subsets of $X$. For each $a \in A$, select $\varepsilon_a$ so that $B_d(a, \varepsilon_a) \cap B = \emptyset$. Similarly, for each $b \in B$, choose $\varepsilon_b$ such that $B_d(b, \varepsilon_b) \cap A = \emptyset$. Define $$U = \bigcup_{a \in A} B_d(a, \varepsilon_a/2), \qquad V = \bigcup_{b \in B} B_d(b, \varepsilon_b/2).$$ By design, $A \subseteq U$ and $B \subseteq V$. We claim that $U \cap V = \emptyset$. If not, there exists $z \in U \cap V$. Then $z \in B_d(a, \varepsilon_a/2) \cap B_d(b, \varepsilon_b/2)$ for some $a \in A$ and $b \in B$. Applying the triangle inequality, we have $$d(a, b) \le d(a, z) + d(z, b) < \tfrac12(\varepsilon_a + \varepsilon_b).$$ If $\varepsilon_a \le \varepsilon_b$, then $a \in B_d(b, \varepsilon_b)$, which is impossible. If $\varepsilon_b \le \varepsilon_a$, then $b \in B_d(a, \varepsilon_a)$, which is also impossible. $\blacksquare$
The halving is the whole trick. Full-size balls might overlap; half-size balls can't, because two half-radii add up to at most the bigger radius. Watch it.
Notice the proof never needs $A$ and $B$ to be a positive distance apart. They can creep toward each other forever, and the balls just get smaller to match. Drag the sets around and see.
The $x$-axis and $y = \frac1x$ get closer forever and never touch, and normality still gives each one its own open set. Healthy boundaries.
The second family trades the metric for compactness. The proof uses compactness in each of two steps: first a point against a closed set, then a closed set against a closed set.
Every compact Hausdorff space is normal.
Proof. We use compactness in each of two steps. One-point sets are closed in a Hausdorff space. Let $A$ and $B$ be disjoint closed subsets of the compact Hausdorff space $X$.
Step 1. Let $x \in A$. For each $b \in B$, there exist disjoint open sets $U_b$ and $V_b$ such that $x \in U_b$ and $b \in V_b$. This is so because $X$ is Hausdorff. Since $B$ is closed and $X$ is compact, $B$ is therefore compact, so the open cover $\{V_b : b \in B\}$ of $B$ has a finite subcover, say $V_{b_1}, \ldots, V_{b_L}$. The sets $$\bigcap_{n=1}^{L} U_{b_n} \quad \text{and} \quad \bigcup_{n=1}^{L} V_{b_n}$$ are open and disjoint. The first contains $x$ and the second is a superset of $B$.
Step 2. For each $a \in A$, by Step 1 there exist disjoint open sets $U_a$ and $V_a$ such that $a \in U_a$ and $B \subseteq V_a$. Since $A$ is compact, we can cover $A$ by finitely many of the sets $U_a$, say $U_{a_1}, \ldots, U_{a_m}$. Then $$\bigcup_{n=1}^{m} U_{a_n} \quad \text{and} \quad \bigcap_{n=1}^{m} V_{a_n}$$ are disjoint open sets that contain $A$ and $B$, respectively. $\blacksquare$
Finitely many is what makes the intersections open. Here are both steps, one after the other.
The third family is where the two halves of this page meet. Regular is weaker than normal, but add a countable basis and it's enough.
Every regular second-countable space is normal.
Proof. Let $X$ be regular with countable basis $\mathcal{B}$, and let $A$, $B$ be disjoint closed sets. One-point sets are closed because $X$ is regular.
Each $a \in A$ has the neighborhood $X \setminus B$. By the lemma, there is a neighborhood $W$ of $a$ with $\overline{W} \subseteq X \setminus B$, and a basis element $U$ with $a \in U \subseteq W$; then $\overline{U}$ misses $B$. The basis elements chosen this way are countably many; list them $U_1, U_2, \ldots$ They cover $A$, and each $\overline{U_n}$ misses $B$. Likewise list basis elements $V_1, V_2, \ldots$ that cover $B$, each $\overline{V_n}$ missing $A$.
The unions $\bigcup U_n$ and $\bigcup V_n$ are open and contain $A$ and $B$, but they may overlap. Shave them: let $$U_n' = U_n \setminus \big(\overline{V_1} \cup \cdots \cup \overline{V_n}\big), \qquad V_n' = V_n \setminus \big(\overline{U_1} \cup \cdots \cup \overline{U_n}\big).$$ Each is open (an open set minus a closed set). Each $\overline{V_k}$ misses $A$, so $U_n'$ still contains $U_n \cap A$, and $U' = \bigcup U_n'$ contains $A$. Likewise $V' = \bigcup V_n'$ contains $B$.
Finally, $U' \cap V' = \emptyset$. If $x \in U_j' \cap V_k'$ with $j \le k$, then $x \in U_j \subseteq \overline{U_j}$, but $V_k'$ has $\overline{U_j}$ removed. The case $k \le j$ is symmetric. $\blacksquare$
Every $U_n$ gets shaved by the closures of the $V$'s that came before it, and the other way around, so whichever of two sets comes later has made room for the earlier one. Watch it on the line.
The proof used the countable basis for exactly one thing: to cut the cover of $A$ down to a list. Lindelöf would do that job too.
Show that every regular Lindelöf space is normal.
Proof. Let $A$, $B$ be disjoint closed sets. First, $A$ is Lindelöf: if $\mathcal{U}$ is a cover of $A$ by open sets of $X$, then $\mathcal{U} \cup \{X \setminus A\}$ is an open cover of $X$, and a countable subcover, minus $X \setminus A$, covers $A$.
By the lemma, each $a \in A$ has a neighborhood $W_a$ with $\overline{W_a}$ missing $B$. The $W_a$ cover $A$, so countably many of them do; list them $U_1, U_2, \ldots$ Likewise get $V_1, V_2, \ldots$ covering $B$ with closures missing $A$. From here the shaving argument of the theorem above runs word for word, since it only used that the $U_n$ and $V_n$ are open, form lists, cover $A$ and $B$, and have closures missing the other set. $\blacksquare$
So $\mathbb{R}_l$, which is regular (in fact normal, below) and Lindelöf, gets normality two ways. Here's the direct proof, which only uses the shape of its basis.
Show that $\mathbb{R}_l$ is normal.
$\mathbb{R}_l$ is finer than $\mathbb{R}$, so one-point sets are closed. Let $A$, $B$ be disjoint closed sets. For each $a \in A$, the open set $\mathbb{R} \setminus B$ contains $a$, so it contains a basis element $[a, x_a)$. For each $b \in B$, likewise choose $[b, x_b)$ missing $A$. Let $U = \bigcup_{a \in A} [a, x_a)$ and $V = \bigcup_{b \in B} [b, x_b)$.
If $[a, x_a)$ met $[b, x_b)$ with $a < b$, then $b \in [a, x_a)$, which misses $B$. If $b < a$, then $a \in [b, x_b)$, which misses $A$. So $U$ and $V$ are disjoint.
Here's the twist. Products of Hausdorff spaces are Hausdorff, and products of regular spaces are regular. Products of normal spaces are not normal, and $\mathbb{R}_l$ is the counterexample: $\mathbb{R}_l \times \mathbb{R}_l$ is regular, but not normal. The proof is this page's Spicy Bonus.
The Sorgenfrey line is perfectly normal on its own. Multiply it by itself and it stops being normal. Honestly, same.
Normal spaces have a superpower: the open sets that separate $A$ from $B$ can be upgraded to a continuous function that is $0$ on $A$ and $1$ on $B$. That's Urysohn's lemma, the next page, and the shrinking lemma above is the tool that builds it.
Every question so far came right after its idea, which is a hint. Down here the hint is gone. First, true or false: a claim about anything on this page, some of them the kind that sounds right and isn't. Decide, then read why.
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Then a proof from this page with one line to fill in. Every wrong line is false or doesn't follow.
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Next, the mixed set. Every question type on this page except true or false goes into one hat, and each click draws a new one at random. All of them need their answer choices, so the switch stays on Choices.
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Last, the Spicy Bonus: the promised proof that $\mathbb{R}_l \times \mathbb{R}_l$ is not normal. It uses separability, a closed set with too many subsets, and one counting argument from set theory.
Let $L = \{(x, -x) : x \in \mathbb{R}\}$, the antidiagonal of $\mathbb{R}_l \times \mathbb{R}_l$.
Proof. (1) $L$ is closed in $\mathbb{R}^2$, and the topology of $\mathbb{R}_l \times \mathbb{R}_l$ is finer, so $L$ is closed there too. For a point $(x, -x)$, the open set $[x, x + 1) \times [-x, -x + 1)$ meets $L$ only in $(x, -x)$: a point $(y, -y)$ in it has $y \ge x$ and $-y \ge -x$, so $y = x$. So every one-point subset of $L$ is open in $L$, and $L$ is discrete. Then every subset $S \subseteq L$ is closed in $L$, and since $L$ is closed, $S$ is closed in $\mathbb{R}_l \times \mathbb{R}_l$.
(2) Suppose $\mathbb{R}_l \times \mathbb{R}_l$ is normal, and let $D = \mathbb{Q} \times \mathbb{Q}$, which is dense, since every basis element $[a, b) \times [c, d)$ contains a point with rational coordinates. For each $S \subseteq L$, the sets $S$ and $L \setminus S$ are disjoint and closed by (1), so choose disjoint open sets $U_S \supseteq S$ and $V_S \supseteq L \setminus S$ (take $U_S = \emptyset$, $V_S$ the whole plane when $S = \emptyset$, and the other way around when $S = L$).
The map $S \mapsto U_S \cap D$ is one-to-one. Let $S \neq T$, say with a point $p \in S \setminus T$. Then $p \in U_S$ and $p \in L \setminus T \subseteq V_T$, so $U_S \cap V_T$ is a nonempty open set, and since $D$ is dense it contains a point $d \in D$. Now $d \in U_S \cap D$, but $d \in V_T$, which is disjoint from $U_T$, so $d \notin U_T \cap D$.
So $\mathcal{P}(L)$ maps one-to-one into $\mathcal{P}(D)$. But $D$ is countable, so $\mathcal{P}(D)$ has the size of $\mathcal{P}(\mathbb{N})$, which is the size of $\mathbb{R}$, which is the size of $L$ (by $x \mapsto (x, -x)$). That gives a one-to-one map from $\mathcal{P}(L)$ into $L$, which Cantor's theorem forbids. So $\mathbb{R}_l \times \mathbb{R}_l$ is not normal. $\blacksquare$