You have two things to show: that if $D$ is closed, then $X$ is Hausdorff, and that if $X$ is Hausdorff, then $D$ is closed.
Suppose first that $D$ is closed in $X\times X$. To show that $X$ is Hausdorff, you must show that if $x$ and $y$ are any two points of $X$, then there are open sets $U$ and $V$ in $X$ such that $x\in U$, $y\in V$, and $U\cap V=\varnothing$. The trick is to look at the point $p=\langle x,y\rangle\in X\times X$. Because $x\ne y$, $p\notin D$. This means that $p$ is in the open set $(X\times X)\setminus D$. Thus, there must be a basic open set $B$ in the product topology such that $p\in B\subseteq(X\times X)\setminus D$. Basic open sets in the product topology are open boxes, i.e., sets of the form $U\times V$, where $U$ and $V$ are open in $X$, so let $B=U\times V$ for such $U,V\subseteq X$. Can you see how to use this to get the desired open neighborhoods of $x$ and $y$?
Now suppose that $X$ is Hausdorff. To show that $D$ is closed in $X\times X$, you need only show that $(X\times X)\setminus D$ is open. To do this, just take any point $p\in(X\times X)\setminus D$ and show that it has an open neighborhood disjoint from $D$. I suggest that you try to reverse what I did above. First, $p=\langle x,y\rangle$ for some $x,y\in X$, and since $p\notin D$, $x\ne y$. Now use disjoint open neighborhoods of $x$ and $y$ to build a basic open box around $p$ that is disjoint from $D$.