The question goes as follows:
Let $(X, \mathcal{T})$ and $(Y, \mathcal{U})$ be topological spaces. Let $A$ and $B$ be subsets of $X$ and $Y$ respectively. Show that $\operatorname{int} (A \times B) = \operatorname{int}(A) \times \operatorname{int}(B)$.
Here's my proof of the above:
The interior of a set $G$ is defined as $$ x \in \operatorname{int}(G) \iff \exists \text{ an open set $U$ containing $x$ such that $U\subseteq G$} $$ Let $(x,y) \in \operatorname{int}(A \times B)$. This means there exists an open set $A_1 \times B_1$ containing $(x,y)$ such that $A_1 \times B_1 \subseteq A \times B \Rightarrow A_1 \subseteq A$ containing $x$ and $B_1 \subseteq B$ containing $y$. Thus $(x, y) \in \operatorname{int}(A) \times \operatorname{int}(B)$.
Conversely, if $(x, y) \in \operatorname{int}(A) \times \operatorname{int}(B)$, then there exists open sets $A_1$ and $B_1$ containing $x$ and $y$ such that $A_1 \subseteq A$ and $B_1 \subseteq B$. So, $A_1 \times B_1$ is a open set in $A \times B$ containing $(x, y)$. Thus, $(x, y) \in \operatorname{int}(A \times B)$.
I am not quite sure about the correctness of my proof. So I thought I would check it. Thanks in advance.