Solution: Since $A$ is closed in $X$ and $B$ is closed in $Y$, so $X\setminus A \left( = A^c \right) $ is open in $X$ and $Y \setminus B \left( = B^c \right) $ is open in $Y$. We need to show that $A \times B$ is closed in $X \times Y$, that is $\left( X\times Y \right) \setminus \left( A \times B \right) $ is open in $X \times Y$. We claim that \[ \left( X\times Y \right) \setminus \left( A \times B \right) = \left( A^c \times Y \right) \cup \left( X \times B^c \right). \] Take $(x,y) \in (X \times Y) \setminus (A \times B)$. Since $(x,y) \notin A \times B$, either $x \notin A$ or $y \notin B$. That is, either $(x,y) \in A^c \times Y$ or $(x,y) \in X \times B^c$, which means $(x,y) \in \left( A^c \times Y \right) \cup \left( X \times B^c \right)$. Therefore, $(X \times Y)\setminus (A \times B) \subseteq \left( A^c \times Y \right) \cup \left( X \times B^c \right)$.

On the other hand, let $(x,y) \in \left( A^c \times Y \right) \cup \left( X \times B^c \right)$. If $(x,y) \in \left( A^c \times Y \right) $, then $x \notin A$, that means, $(x,y) \in X \times Y$ but $(x,y) \notin A \times B$. Therefore, $(x,y) \in (X \times Y) \setminus (A \times B)$. Hence, $A^c \times Y \subseteq (X \times Y) \setminus (A \times B)$. Similarly, $X \times B^c \subseteq (X \times Y) \setminus (A \times B)$, and hence, $\left( A^c \times Y \right) \cup \left( X \times B^c \right) \subseteq (X \times Y) \setminus (A \times B)$.

Since $A^c$ is open in $X$, and product of open sets is open implies that $A^c \times Y$ is open in $X \times Y$. Similarly, $X \times B^c$ is open in $X \times Y$ and hence their union is open is $X \times Y$. Therefore, $A \times B$ is closed in $X \times Y$.