[Proof Verification]: $\overline{A\times B}=\overline{A}\times\overline{B}$ Suppose $X$ and $Y$ are topological spaces, and $A\subset X$ and $B\subset Y$ are subspaces. I'm trying to prove $\overline{A\times B}=\overline{A}\times\overline{B}$. For the backward inclusion, choose $(x_1,x_2)\in \overline{A}\times \overline{B}$. Every neighbourhood of $x_1$ intersects $A$, and every neighbourhood of $x_2$ intersects $B$, so clearly every neighbourhood of $(x_1,x_2)$ intersects $A\times B$. Conversely, take a point $(x_1,x_2)\in \overline{A\times B}$ and for a contradiction, suppose $(x_1,x_2)\not\in\overline{A}\times\overline{B}$. Then every neighbourhood of $(x_1,x_2)$ intersects $A\times B$. However since neighbourhoods of $x_1$ and $x_2$ which do not intersect $A$ and $B$ respectfully exist, call them $U$ and $V$, the product $U\times V$ does not intersect $A\times B$.$\hspace{1em}\Box$
 A: Here is another way to look at this.
Since
$A\times B= (A\times Y)\cap(X\times B)$,
$$\overline{A\times B}\subset \overline{A\times Y}\cap \overline{X\times B}\subset  (\overline{A}\times Y) \cap (X\times \overline{B}) = \overline{A}\times\overline{B}$$
Here we have used the fact that $\overline{U\cap V}\subset \overline{U}\cap \overline{V}$ for any subsets in a topological space, along with the facts that $A\times Y\subset \overline{A}\times Y$, and $\overline{A}\times Y$ is closed in $X\times Y$. (similarly argument for $X\times B$.)
It remains to show that $\overline{A}\times\overline{B}\subset\overline{A\times B}$. Let $(x,y)\in\overline{A}\times\overline{B}$. Any neighborhood $W$ in $X\times Y$ of $(x,y)$ contains an open set of the form $U\times V$ where $x\in U$, $y\in V$ and $U$, $V$ one in $X$ and $Y$ respectively. Then
$$W\cap(A\times B)\supset (U\times V)\cap (A\times B)=(U\cap A)\times (V\cap B)\neq\emptyset$$
for $x\in\overline{A}$ and $y\in\overline{B}$. Consequently, $(x,y)\in \overline{A\times B}$
