A bare topological space can be very weird: points that can't be told apart by open sets, sequences that converge to every point at once. Most spaces you actually care about avoid this by satisfying one extra condition, named after Felix Hausdorff. It's the first of several "niceness" conditions you'll see in topology, and by far the most used.
The condition asks that any two different points can be put in non-overlapping open sets.
A topological space $(X, \mathcal{T})$ is called a Hausdorff space if for each pair of distinct points $x_1, x_2 \in X$, there exist neighborhoods $U_1$ and $U_2$ of $x_1$ and $x_2$, respectively, that are disjoint.
Hausdorff spaces are also called $T_2$ spaces. There's a whole ladder: $T_0$, $T_1$, $T_2$, $T_{2\frac{1}{2}}$, $T_3$, $T_{3\frac{1}{2}}$, $T_4$... Yes, "$T$ three-and-a-half" is a real thing people say with a straight face.
The real line is the model example.
Show that $\mathbb{R}$ (standard topology) is Hausdorff.
Let $x_1 \neq x_2$ and set $r = \frac{|x_1 - x_2|}{2} > 0$. Then $U_1 = (x_1 - r, x_1 + r)$ and $U_2 = (x_2 - r, x_2 + r)$ are neighborhoods of $x_1$ and $x_2$, and they are disjoint: if $z$ were in both, then $|x_1 - x_2| \le |x_1 - z| + |z - x_2| < r + r = |x_1 - x_2|$, which is impossible.
Two quick examples at the extremes, using the topologies from the first page.
Which of these are Hausdorff: a set with the discrete topology, and a set with at least two points and the trivial topology?
Discrete: yes. $\{x_1\}$ and $\{x_2\}$ are open and disjoint.
Trivial: no. The only neighborhood of any point is the whole space $X$, and $X \cap X \neq \emptyset$.
The first payoff: in a Hausdorff space, points are closed.
Every finite point set in a Hausdorff space $X$ is closed.
Proof. First we show a one-point set $\{x_0\}$ is closed. If $x$ is a point of $X$ other than $x_0$, then since $X$ is Hausdorff there are disjoint neighborhoods $U$ and $V$ of $x$ and $x_0$, respectively. Since $U$ does not intersect $\{x_0\}$, $x \notin \overline{\{x_0\}}$. Therefore $\overline{\{x_0\}} = \{x_0\}$, so $\{x_0\}$ is closed. A finite point set is a finite union of one-point sets, and finite unions of closed sets are closed. $\blacksquare$
We used the test from Closed Sets: $x \in \overline{A}$ exactly when every neighborhood of $x$ intersects $A$. One neighborhood that misses $A$ is enough to kick $x$ out of the closure.
Being Hausdorff survives taking products: two points of $X \times Y$ that differ must differ in some coordinate, and we separate them there.
The product of two Hausdorff spaces is Hausdorff.
Proof. Let $X$ and $Y$ be Hausdorff, and let $(a, b)$ and $(p, q)$ be distinct points of $X \times Y$. If $a \neq p$, there are disjoint open sets $U$ and $V$ of $X$ with $a \in U$ and $p \in V$. Then $U \times Y$ and $V \times Y$ are open in $X \times Y$, contain $(a, b)$ and $(p, q)$ respectively, and $(U \times Y) \cap (V \times Y) = (U \cap V) \times Y = \emptyset$.
Otherwise $a = p$, so $b \neq q$, and there are disjoint open sets $W$ and $\tilde{W}$ of $Y$ with $b \in W$ and $q \in \tilde{W}$. Then $X \times W$ and $X \times \tilde{W}$ are disjoint open sets containing $(a, b)$ and $(p, q)$. Either way the points are separated, so $X \times Y$ is Hausdorff. $\blacksquare$
(The Product Topology page extends this to products of any number of Hausdorff spaces.) There's a slick way to say "Hausdorff" using the product $X \times X$. It needs a name for the line $y = x$.
Let $X$ be a set. The diagonal of $X$ is the set $\Delta = \{(x, x) : x \in X\} \subset X \times X$.
A point $(x, y)$ off the diagonal is a pair of distinct points. Separating $x$ and $y$ by disjoint open sets $U$ and $V$ is the same as finding an open box $U \times V$ around $(x, y)$ that misses $\Delta$ (a point $(z, z)$ in $U \times V$ would put $z$ in $U \cap V$). Drag the point and see.
So "$\Delta$ has an open complement" and "points can be separated" are the same statement. Proving that carefully is Exercise 1.
On the Closed Sets page we saw a sequence converge to every real number at once. Hausdorff spaces don't allow that.
If $(X, \mathcal{T})$ is a Hausdorff space, then a sequence of points of $X$ converges to at most one point of $X$.
Proof. Suppose $(x_n)_{n=1}^\infty$ converges to $x \in X$, and let $y \neq x$. Since $X$ is Hausdorff, there are disjoint neighborhoods $U$ and $V$ of $x$ and $y$. Since $U$ contains $x_n$ for all but finitely many $n$, the disjoint set $V$ can contain $x_n$ for only finitely many $n$. So $(x_n)$ does not converge to $y$. $\blacksquare$
Compare the two topologies on $\mathbb{R}$ directly. In the standard topology, try to make $\frac{1}{n}$ converge to something other than $0$. In the finite complement topology, try to stop it from converging to anything.
The finite complement topology can't separate points, so the theorem doesn't protect it. Exercise 2 makes that official.
The converse of the theorem is false for sequences: there are non-Hausdorff spaces where every sequence has at most one limit. Sequences just can't see enough of a general space. If you want a test that works both ways, you need nets.
On the Nets page we'll prove that $X$ is Hausdorff if and only if every convergent net has exactly one limit.
The first one is the diagonal criterion from the widget.
Show that $(X, \mathcal{T})$ is Hausdorff if and only if its diagonal $\Delta$ is closed in $X \times X$.
Proof. Suppose $X$ is Hausdorff. Let $(x, y) \in (X \times X) \setminus \Delta$. Then $x \neq y$, so there are disjoint neighborhoods $U$ and $V$ of $x$ and $y$. We have $(x, y) \in U \times V \subset (X \times X) \setminus \Delta$, because if $(z, z) \in U \times V$ then $z \in U \cap V = \emptyset$. So every point of the complement of $\Delta$ has a neighborhood inside the complement; the complement is open and $\Delta$ is closed.
Conversely, suppose $\Delta$ is closed, and let $p, q \in X$ be distinct. Then $(p, q) \notin \Delta$, and the complement of $\Delta$ is open, so there is a basis element $U \times V$ with $(p, q) \in U \times V \subset (X \times X) \setminus \Delta$. Then $p \in U$, $q \in V$, and $U \cap V = \emptyset$ (a point $z$ in both would give $(z, z) \in U \times V$). So $X$ is Hausdorff. $\blacksquare$
Now prove what the widget showed.
Give $\mathbb{R}$ the finite complement topology. Is $(\mathbb{R}, \mathcal{T}_f)$ a Hausdorff space?
No. Suppose $U$ and $V$ are open sets containing distinct points $x$ and $y$. Both are nonempty, so in $\mathcal{T}_f$ their complements are finite: say $$\mathbb{R} \setminus U = \{p_1, \dots, p_n\} \quad \text{and} \quad \mathbb{R} \setminus V = \{q_1, \dots, q_m\}.$$ Then $\mathbb{R} \setminus (U \cap V) = (\mathbb{R} \setminus U) \cup (\mathbb{R} \setminus V)$ is finite, and since $\mathbb{R}$ is infinite there is a point $z$ different from every $p_i$ and every $q_j$. That $z$ lies in $U \cap V$, so $U$ and $V$ are not disjoint. No two points can be separated, so $(\mathbb{R}, \mathcal{T}_f)$ is not Hausdorff.
The next one is a warm-up for the Spicy Bonus. Read the hypotheses carefully.
Let $X$ be a topological space and $A, B \subset X$ with $X = A \cup B$. If $A$ and $B$ are Hausdorff (as subspaces), is $X$ Hausdorff?
No. The minimalist counterexample: $A = \{0\}$, $B = \{1\}$, and $X = A \cup B = \{0, 1\}$ with the trivial topology. Each of $A$ and $B$ has only one point, so they are trivially Hausdorff. But the only neighborhood of $0$ or $1$ in $X$ is $X$ itself, so $0$ and $1$ cannot be separated, and $X$ is not Hausdorff.
What went wrong there is that $A$ and $B$ weren't closed. With that hypothesis added, the answer flips, but the proof gets much harder.
In "is $X$ Hausdorff?" problems with $X = A \cup B$, split into cases by where the two points live: both in $X \setminus A$, both in $X \setminus B$, one in each, or at least one in $A \cap B$. Closed sets make $X \setminus A$ and $X \setminus B$ open, which is what lets you trim the separating sets so they stay open in $X$.
Indeed this question is far harder than the version without $A$ and $B$ being closed.
Let $X$ be a topological space and $A, B \subset X$ be closed subsets of $X$ such that $X = A \cup B$. If $A$ and $B$ are Hausdorff, is $X$ Hausdorff?
Yes. Let $p \neq q$ in $X$. Since $A$ and $B$ are closed, $X \setminus A$ and $X \setminus B$ are open in $X$. Also $X \setminus B \subset A$ and $X \setminus A \subset B$, because $X = A \cup B$. Recall that "open in $A$" means "of the form $A \cap U$ with $U$ open in $X$," so separating $p$ and $q$ inside $A$ gives open sets $U, V$ of $X$ with $p \in U$, $q \in V$, and $A \cap U \cap V = \emptyset$.
Case 1: $p, q \in A \cap B$. Separate them in $A$ by $U_A, V_A$ (so $A \cap U_A \cap V_A = \emptyset$) and in $B$ by $U_B, V_B$ (so $B \cap U_B \cap V_B = \emptyset$). Then $U = U_A \cap U_B$ and $V = V_A \cap V_B$ are open in $X$ and contain $p$ and $q$. A point $z \in U \cap V$ lies in $A$ or in $B$; either way it lies in a set we chose to be empty. So $U \cap V = \emptyset$.
Case 2: $p, q \in A$, and at least one of them is in $X \setminus B$. Separate them in $A$ by $U, V$ with $A \cap U \cap V = \emptyset$. Cut each of $U$, $V$ whose point lies in $X \setminus B$ down by intersecting it with $X \setminus B$; the new sets are still open and still contain $p$ and $q$. A common point $z$ of the new sets would lie in $X \setminus B \subset A$ and in $U \cap V$, so $z \in A \cap U \cap V = \emptyset$. So they are disjoint.
Case 3: $p, q \in B$, and at least one of them is in $X \setminus A$. Same as Case 2 with $A$ and $B$ swapped.
Case 4: one point is in $X \setminus B$ and the other is in $X \setminus A$. These two sets are open and disjoint, since $(X \setminus B) \cap (X \setminus A) = X \setminus (A \cup B) = \emptyset$.
Every pair is covered. If both points are in $A \cap B$, use Case 1. Otherwise one of them, say $p$, is outside $A \cap B$; by symmetry say $p \in X \setminus B$ (so $p \in A$). If $q \in A$, use Case 2; if not, $q \in X \setminus A$ and we use Case 4. So $X$ is Hausdorff. $\blacksquare$