Let $(X,\tau)$ be a topological space. Show that $int(A) \cap int(B)$ $\subset$ $int(A \cap B)$
Proof: x $\in$ $int(A)$ and $x\in int(B)$ means that $\exists$ $U_1 (open)$ containing x so that $U_1 \subset A$ and $\exists U_2$ open containing x so that $U_2 \subset B$ hence $U_1 \cap U_2$ is open and a subset of $A \cap B$. Since x is contained in $U_1$ and $U_2$ it follows that x is in the interior of $A\cap B$.
Is this proof correct?
Yes, correct.
Alternative way that makes use of the following characterization of "interior":
(If this characterization is not yet familiar to you then I advice you to make it familiar to you)
You could also say: $\mathsf{int}(C)$ is the largest open subset of $C$. This in the sense that $\mathsf{int}(C)$ is an open subset of $C$, and secondly that every open subset of $C$ is a subset of $\mathsf{int}(C)$.
Proof:
We have $\mathsf{int}(A)\subseteq A$ and $\mathsf{int}(B)\subseteq B$ by definition and consequently $\mathsf{int}(A)\cap\mathsf{int}(B)\subseteq A\cap B$.
As a finite intersection of open sets the set $\mathsf{int}(A)\cap\mathsf{int}(B)$ is an open set, hence must be a subset of $\mathsf{int}(A\cap B)$.