Once you have a topology on $X$, every subset of $X$ gets one for free. That's the first of the "new spaces from old" constructions (products and quotients come later), and it's the one you'll use without thinking more than any other. The catch is that "open" now depends on which space you're standing in.
The rule is the simplest thing that could work: cut every open set of $X$ down to $Y$.
Let $X$ be a topological space with topology $\mathcal{T}$. If $Y$ is a subset of $X$, the collection $$\mathcal{T}_Y = \{Y \cap U : U \in \mathcal{T}\}$$ is a topology on $Y$, called the subspace topology.
(It really is a topology: $\emptyset = Y \cap \emptyset$ and $Y = Y \cap X$, and unions and finite intersections pass straight through the "$Y \cap$" part. For example, $(Y \cap U_1) \cap (Y \cap U_2) = Y \cap (U_1 \cap U_2)$.) A subset that carries this topology gets a name.
With the subspace topology, $Y$ is called a subspace of $X$. Its open sets consist of all intersections of open sets of $X$ with $Y$.
Here's the example that surprises everyone the first time.
Let $Y = [0, 1]$, as a subspace of $\mathbb{R}$. Is $[0, \frac{1}{2})$ open in $Y$?
Yes. $[0, \frac{1}{2}) = Y \cap (-1, \frac{1}{2})$, and $(-1, \frac{1}{2})$ is open in $\mathbb{R}$. So $[0, \frac{1}{2})$ is open in $Y$, even though it is not open in $\mathbb{R}$ (no open interval around $0$ fits inside it).
So $[0, \frac{1}{2})$ is open now? I spent all of calculus being told half-open intervals aren't open. Betrayal.
The point is that inside $Y$ there's nothing to the left of $0$ for an interval to poke into. Drag an open interval of $\mathbb{R}$ across a few different subspaces and watch what's left.
In $\mathbb{Z}$, every single point is open: $\{n\} = \mathbb{Z} \cap (n - \frac{1}{2}, n + \frac{1}{2})$. So $\mathbb{Z}$, as a subspace of $\mathbb{R}$, has the discrete topology.
Never write "$A$ is open" when there are two spaces around. Write "$A$ is open in $Y$" or "$A$ is open in $X$." The two can disagree, and a proof that mixes them up is wrong even when it looks right.
A basis for $X$ cuts down to a basis for $Y$, the same way open sets do.
If $\mathcal{B}$ is a basis for the topology of $X$, then the collection $$\mathcal{B}_Y = \{B \cap Y : B \in \mathcal{B}\}$$ is a basis for the subspace topology on $Y$.
Proof. Each $B \cap Y$ is open in $Y$. Given $U$ open in $X$ and $y \in U \cap Y$, choose $B \in \mathcal{B}$ with $y \in B \subset U$. Then $y \in B \cap Y \subset U \cap Y$. By the theorem on recognizing a basis (from Topological Spaces & Bases), $\mathcal{B}_Y$ is a basis for the subspace topology on $Y$. $\blacksquare$
When is "open in $Y$" automatically "open in $X$"? When $Y$ itself is open.
Let $Y$ be a subspace of $X$. If $U$ is open in $Y$ and $Y$ is open in $X$, then $U$ is open in $X$.
Proof. Since $U$ is open in $Y$, $U = Y \cap V$ for some $V$ open in $X$. Since $Y$ and $V$ are both open in $X$, so is $Y \cap V$. $\blacksquare$
That explains the example: $[0, 1]$ is not open in $\mathbb{R}$, so its open sets don't have to be open in $\mathbb{R}$. Try it on a four-point space, where you can see every open set at once. The topology on $X = \{a, b, c, d\}$ is $\{\emptyset, \{a\}, \{c\}, \{a, b\}, \{a, c\}, \{a, b, c\}, X\}$.
Notice that different open sets of $X$ can have the same trace in $Y$, so the subspace usually has fewer open sets. And whenever you pick a $Y$ that is in the topology, the orange chips disappear, just like the theorem says.
On the first page we built topologies from scratch. Subspaces are the first time a topology arrives already built, inherited from a bigger space.
If $A \subset X$ and $B \subset Y$, there are two reasonable ways to put a topology on $A \times B$: take the product of the subspace topologies, or take the subspace topology from $X \times Y$. Good news: they agree.
If $A$ is a subspace of $X$ and $B$ is a subspace of $Y$, then the product topology on $A \times B$ is the same as the topology $A \times B$ inherits as a subspace of $X \times Y$.
Proof. Let $U \times V$ be a general basis element of $X \times Y$, where $U$ is open in $X$ and $V$ is open in $Y$. By the theorem above, $(U \times V) \cap (A \times B)$ is a general basis element for the subspace topology on $A \times B$. Now $$(U \times V) \cap (A \times B) = (U \cap A) \times (V \cap B).$$ Since $U \cap A$ and $V \cap B$ are the general open sets of the subspaces $A$ and $B$, the set $(U \cap A) \times (V \cap B)$ is the general basis element for the product topology on $A \times B$. So the two topologies have the same basis, and therefore they are the same topology. $\blacksquare$
We'll define the product topology properly later. For now, "basis = open rectangles $U \times V$" is all this theorem uses.
So the unit square $[0, 1] \times [0, 1]$ has only one sensible topology, whichever way you build it.
Is $[0, \frac{1}{2}) \times (\frac{1}{4}, \frac{3}{4})$ open in the unit square $I \times I$, where $I = [0, 1]$?
Yes. $[0, \frac{1}{2})$ is open in $I$ and $(\frac{1}{4}, \frac{3}{4})$ is open in $I$, so their product is a basis element of the product topology on $I \times I$. Equivalently, it equals $\big((-1, \frac{1}{2}) \times (\frac{1}{4}, \frac{3}{4})\big) \cap (I \times I)$, an open set of $\mathbb{R}^2$ intersected with the square.
Start with a finite one you can do by listing.
Let $X = \{1, 2, 3, 4\}$ have the topology $\mathcal{T} = \{\emptyset, X, \{1\}, \{1, 2\}, \{1, 3\}, \{1, 2, 3\}\}$. Let $A = \{1, 2, 3\} \subset X$ have the subspace topology. Find all subsets $B \subset A$ such that $B \notin \mathcal{T}$.
Intersecting $A$ with each $T \in \mathcal{T}$, the open sets of $A$ are $\mathcal{T}_A = \{\emptyset, A, \{1\}, \{1, 2\}, \{1, 3\}\}$. The subsets of $A$ that are not in $\mathcal{T}$ are $\{2\}$, $\{3\}$, and $\{2, 3\}$. (Their complements in $A$ are $\{1, 3\}$, $\{1, 2\}$, and $\{1\}$, which are open in $A$, so these three sets are the nontrivial closed sets of $A$. More on that on the next page.)
Next, describe all the basis elements of a subspace of $\mathbb{R}$.
Consider the subset $I = [0, 1]$ of $\mathbb{R}$, in the subspace topology. What are the forms of the basis elements $(a, b) \cap I$ of $\mathcal{T}_I$?
The subspace topology has as a basis all sets of the form $(a, b) \cap I$, where $(a, b)$ is an open interval in $\mathbb{R}$. These sets have the following forms: $$(a, b) \cap I = \begin{cases} (a, b) & \text{if } a \text{ and } b \text{ are in } I, \\ [0, b) & \text{if only } b \text{ is in } I, \\ (a, 1] & \text{if only } a \text{ is in } I, \\ I \text{ or } \emptyset & \text{if neither } a \text{ nor } b \text{ is in } I. \end{cases}$$ Notice that the open sets of the subspace are not always open in the larger space.
This one is a classic. Be careful with the absolute values: each set has two halves.
Consider the set $Y = [-1, 1]$ as a subspace of $\mathbb{R}$. Which of the following sets are open in $Y$? Which are open in $\mathbb{R}$?
Watch out for a classic slip in that last exercise.
"Open in $Y$ and a subset of $Y$" does not mean "open in $\mathbb{R}$." Part (b) is the trap: $B$ looks like a union of intervals, but its closed ends $\pm 1$ are only allowed because $Y$ stops there.
One more sanity check on the definition: taking a subspace of a subspace shouldn't give anything new.
Show that if $Y$ is a subspace of $X$ and $A$ is a subset of $Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$.
Proof. Write $\mathcal{T}_{A,Y}$ and $\mathcal{T}_{A,X}$ for the subspace topologies on $A$ induced by $Y$ and by $X$. We show $\mathcal{T}_{A,Y} = \mathcal{T}_{A,X}$.
Let $U \in \mathcal{T}_{A,Y}$. Then $U = A \cap V$, where $V$ is open in $Y$. Because $Y$ is a subspace of $X$, there is $W$ open in $X$ with $V = Y \cap W$. So $U = A \cap (Y \cap W) = (A \cap Y) \cap W = A \cap W$, since $A \subset Y$. Hence $U \in \mathcal{T}_{A,X}$.
Conversely, let $\tilde{U} \in \mathcal{T}_{A,X}$, so $\tilde{U} = A \cap \tilde{W}$ for some $\tilde{W}$ open in $X$. Then $\tilde{U} = (A \cap Y) \cap \tilde{W} = A \cap (Y \cap \tilde{W})$. Since $Y \cap \tilde{W}$ is open in $Y$, we conclude $\tilde{U} \in \mathcal{T}_{A,Y}$. Therefore $\mathcal{T}_{A,Y} = \mathcal{T}_{A,X}$. $\blacksquare$