Is product of two closed sets closed? If $A$ is closed in $X$ and $B$ is closed in $Y$, 

does it follow that $A \times B$ is closed in $X \times Y$ ?
$A$ is closed in $X$, so $A^c$ is open, likewise for $B$ in $Y$.

Moreover, $A^c \times Y$ and $X \times B^c$ are both open in $X \times Y$. 

Thus $$(A \times B)^c = (A^c \times Y) \cup (X \times B^c)  $$ is open. Hence $A \times B$ is closed.
Given $A$ and $B$ are closed sets, let $(x,y)$ be a limit point of $A \times B$. Then there exists a sequence $\{(x_n, y_n)\}$ in $A \times B$ such that $\{(x_n, y_n)\}$ converges to $(x, y)$.
Since $\{(x_n, y_n)\}$ is a sequence in $A \times B$, $\{x_n\}$ is a sequence in $A$ and $\{y_n\}$ is a sequence in $B$.
Also, since $\{(x_n, y_n)\}$ converges to $(x, y)$, $\{x_n\}$ converges to $x$ and $\{y_n\}$ converges to $y$.
We obtain a sequence $\{x_n\}$ in $A$ converging to $x$. This implies that $x$ is a limit point of $A$ and hence it is in the closure of $A$. Since $A$ is closed, $A$ equals it closure, i.e. $x$ is in $A$.
Similarly, we can prove $y$ is in $B$.
Therefore, $(x, y)$ is in $A \times B$, i.e. $A \times B$ contains all its limit points.
Hence $A \times B$ is closed.