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?