[proofplan]
We establish the characterisation of $\operatorname{int}_A(B)$ by unwinding the definition of open sets in the subspace topology. Every open set in $A$ has the form $U \cap A$ for some $U \in \tau$, so a point lies in $\operatorname{int}_A(B)$ precisely when it belongs to some $U \cap A \subset B$. For the inclusion $\operatorname{int}_X(B) \cap A \subset \operatorname{int}_A(B)$, we observe that any ambient open set witnessing membership in $\operatorname{int}_X(B)$ also witnesses membership in $\operatorname{int}_A(B)$ after intersecting with $A$. A concrete example then shows the reverse inclusion can be strict.
[/proofplan]
[step:Characterise $\operatorname{int}_A(B)$ via the subspace topology]
By definition, $\operatorname{int}_A(B)$ is the largest set that is open in $A$ and contained in $B$. A set $V$ is open in the subspace topology $\tau_A$ if and only if $V = U \cap A$ for some $U \in \tau$. Therefore $x \in \operatorname{int}_A(B)$ if and only if there exists $U \in \tau$ with $x \in U \cap A \subset B$. Since $B \subset A$, the condition $U \cap A \subset B$ is equivalent to requiring that $U \cap A \subset B$, which is the set displayed in the theorem statement.
[guided]
We need to connect the abstract definition of interior — the largest open subset — to the concrete characterisation in the statement. In the subspace topology $\tau_A = \{U \cap A : U \in \tau\}$, a set $V \subset A$ is open in $A$ if and only if $V = U \cap A$ for some $U \in \tau$. The interior $\operatorname{int}_A(B)$ is the union of all open-in-$A$ sets contained in $B$:
\begin{align*}
\operatorname{int}_A(B) = \bigcup \{ U \cap A : U \in \tau, \; U \cap A \subset B \}.
\end{align*}
A point $x$ belongs to this union if and only if there exists at least one $U \in \tau$ with $x \in U \cap A \subset B$. This is precisely the set-builder description in the theorem statement.
[/guided]
[/step]
[step:Prove $\operatorname{int}_X(B) \cap A \subset \operatorname{int}_A(B)$]
Let $x \in \operatorname{int}_X(B) \cap A$. Since $x \in \operatorname{int}_X(B)$, there exists $U \in \tau$ with $x \in U \subset B$. Then $x \in U \cap A$ and $U \cap A \subset U \subset B$, so $x \in \operatorname{int}_A(B)$ by the characterisation established above.
[guided]
Take any $x \in \operatorname{int}_X(B) \cap A$. The membership $x \in \operatorname{int}_X(B)$ provides an ambient open set $U \in \tau$ with $x \in U \subset B$. We need to show $x \in \operatorname{int}_A(B)$, i.e., we need an ambient open set $W \in \tau$ with $x \in W \cap A \subset B$. The same set $U$ works: since $x \in A$ by hypothesis, we have $x \in U \cap A$, and since $U \subset B$, the intersection $U \cap A \subset U \subset B$. Hence $x \in \operatorname{int}_A(B)$.
The key point is that an ambient open neighbourhood witnessing $x \in \operatorname{int}_X(B)$ automatically serves as a witness for $x \in \operatorname{int}_A(B)$ — intersecting with $A$ can only shrink the neighbourhood, and it was already contained in $B$.
[/guided]
[/step]
[step:Exhibit a case where the inclusion is strict]
Consider $X = \mathbb{R}$ with the standard topology, $A = [0, 1]$ with the subspace topology, and $B = [0, 1)$. We claim $\operatorname{int}_X(B) \cap A \subsetneq \operatorname{int}_A(B)$.
The point $0$ does not belong to $\operatorname{int}_X(B)$: every open set $U$ in $\mathbb{R}$ containing $0$ contains points $x < 0$, so $U \not\subset B = [0, 1)$. Hence $0 \notin \operatorname{int}_X(B) \cap A$.
However, $0 \in \operatorname{int}_A(B)$: take $U = (-1, 1) \in \tau$. Then $0 \in U \cap A = (-1, 1) \cap [0, 1] = [0, 1) = B$, so $U \cap A \subset B$. Hence $0 \in \operatorname{int}_A(B)$.
Therefore $\operatorname{int}_X(B) \cap A \subsetneq \operatorname{int}_A(B)$, confirming that the inclusion can be strict.
[guided]
To see that equality can fail, we construct a concrete example. Let $X = \mathbb{R}$, let $A = [0, 1]$ carry the subspace topology, and let $B = [0, 1)$.
First, $\operatorname{int}_X(B) = (0, 1)$: any open interval around $0$ in $\mathbb{R}$ extends below $0$ and thus exits $B = [0, 1)$, and any open interval around a point in $[1, \infty)$ exits $B$ as well. So $\operatorname{int}_X(B) \cap A = (0, 1) \cap [0, 1] = (0, 1)$.
Now consider $\operatorname{int}_A(B)$. The point $0$ satisfies $0 \in A$ and $0 \in B$. Choose $U = (-1, 1) \in \tau$. Then $U \cap A = (-1, 1) \cap [0, 1] = [0, 1) = B$, so $0 \in U \cap A \subset B$, confirming $0 \in \operatorname{int}_A(B)$.
Why does the discrepancy occur? In the ambient space $\mathbb{R}$, every open neighbourhood of $0$ must extend to the left of $0$, exiting $B = [0, 1)$. But in the subspace $A = [0, 1]$, points to the left of $0$ do not exist — the intersection with $A$ removes them, leaving $[0, 1)$ open in $A$. This is the fundamental mechanism: the subspace topology can create "new" open sets that are not open in the ambient space.
We conclude $\operatorname{int}_A(B) = [0, 1) \supsetneq (0, 1) = \operatorname{int}_X(B) \cap A$.
[/guided]
[/step]
