Let $X,Y$ be topological spaces. Prove that $\operatorname{cl}(A \times B) = \operatorname{cl}_X(A) \times \operatorname{cl}_Y(B)$ in the product topology.
My attempt:
Denote $\mathcal{V(x,y)}$ for the set of all neighborhoods of $(x,y)$
$"\subseteq"$ Let $(x,y) \in \operatorname{cl}(A \times B)$. Then, for all $V \times W \in \mathcal{V}(x,y)$, we have that $(V \times W) \cap (A \times B) \neq \emptyset$. Since $\mathcal{V(x)} \times \mathcal{V}(y) \subseteq\mathcal{V}(x,y)$, it follows that for $V \in \mathcal{V}(x), W \in \mathcal{V}(y)$
$(V \cap A) \times (W \cap B) = (V \times W) \cap (A \times B) \neq \emptyset$
Hence, $V \cap A \neq \emptyset \neq W \cap B$ and it follows that $x \in \operatorname{cl}_X(A), y \in \operatorname{cl}_Y(B)$
$"\supseteq"$ Same argument, where we notice that $\mathcal{V}(x) \times \mathcal{V}(y)$ is a basis for the neighborhoods $\mathcal{V}(x,y)$, and if it is true for every element in a basis, then it is true for every neighborhood.
Is this correct?
$\subseteq$ correct. This is another route:
$\overline{A}\times\overline{B}$ is closed and contains $A\times B$ as a subset, so that:$$\overline{A\times B}\subseteq\overline{A}\times\overline{B}$$
$\supseteq$ correct. A route that does not mention "basis of neighborhoods":
Let it be that $x\in\overline A$ and $y\in\overline B$ and let $U\subseteq X\times Y$ be an open set with $\langle x,y\rangle\in U$.
Then open sets $V\subseteq X$ and $W\subseteq Y$ exist with $\langle x,y\rangle\in V\times W\subseteq U$. Then $V\cap A\neq\varnothing$ and $W\cap B\neq\varnothing$ and if $a\in V\cap A$ and $b\in W\cap B$ then $\langle a,b\rangle\in U\cap(A\times B)$.
This tells us that $\langle x,y\rangle\in\overline{A\times B}$ and proved is now that:$$\overline{A}\times\overline{B}\subseteq\overline{A\times B}$$
If there is indeed familiarity with neighborhood bases then your proof is shorter.