Solution: Let us suppose that $X$ be Hausdorff. We will show that the complement of the diagonal is open in $X \times X$. Define \[ U \coloneqq \Delta ^c = \left\{ (x,y) \in X \times X : x \neq y \right\} . \] Since $X$ is Hausdorff, for any $(x,y) \in U$ we can find disjoint open neighborhoods $V_x$ and $V_y$ of $x$ and $y$ respectively. Since $V_x \cap V_y = \emptyset $, so $(V_x \times V_y) \cap (X \times X) = \emptyset $. Therefore, $V_x \times V_y \subseteq U$ and hence, $U$ is open.

Conversely, assume that $\Delta $ is closed. Let $x \neq y$ be two points in $X$. This implies that $(x,y) \in \Delta ^c$, which is an open set. Hence, there exists basis set for the product topology $U \times V \subseteq X \times X$ satisfying $(x,y) \in U \times V \subseteq \Delta ^c$. Thus, $x\in U$ and $y \in V$ and $U \cap V = \emptyset $. Therefore, $X$ is Hausdorff.