I'm trying to prove the following.
Show that $X$ is Hausdorff if and only if the diagonal $\Delta = \{x \times x \mid x \in X\}$ is closed in $X \times X$.
Here is my attempt.
($\Leftarrow$) Suppose that $\Delta$ is closed in $X \times X$, and let $x \neq y$ in $X$. Then $(x,y) \not \in \Delta$. In particular, as $\Delta$ is closed, $X \times X - \Omega$ is open and contains $(x,y)$, so there exists a basis element, $U \times V$, for the product topology on $X \times X$ such that $(x,y) \in U \times V \subset X \times X - \Delta$. Then $U$ and $V$ are neighborhoods in $X$ which contain $x$ and $y$, respectively. It suffices to prove that $U \cap V = \emptyset$. Indeed, if $m \in U \cap V$, then $(m,m) \in \Delta \cap U \times V$. But $U \times V \subset X \times X - \Delta$, implying that $U \times V$ and $\Delta$ are disjoint, so we have a contradiction. Therefore, $U \cap V = \emptyset$ and $X$ is Hausdorff.
($\Rightarrow$) Suppose that $X$ is Hausdorff. To show that $\Delta$ is closed, we will show that $X \times X - \Delta$ is open in the product topology on $X \times X$, which is true if and only if for every $p \in X \times X - \Delta$, there exists a basis element $U \times V$ such that $p \in U \times V \subset X \times X - \Delta$. Let $x \neq y$ in $X$. As $X$ is Hausdorff, there exist disjoint neighborhoods $U$ and $V$ in $X$ which contain $x$ and $y$, respectively. Then $U \times V$ is a basis element for the product topology which contains the point $(x,y)$. It suffices to show that $U \times V \cap \Delta = \emptyset$. If there exists $(m,n) \in U \times V \cap \Delta$, then we have $m \in U$, $n \in V$, and $m=n \in U \cap V$, contradicting disjointness of $U$ and $V$. so $U \times V \cap \Delta = \emptyset$, so $(x,y) \in U \times V \subset X \times X - \Delta$, so $X \times X - \Delta$ is open, and hence $\Delta$ is closed.
How does this look?
