Theorem 1   A space $X$ is Hausdorff if and only if $$ \{ (x,x)\in X\times X \mid x\in X\}$$ is closed in $X\times X$ under the product topology.

Proof. First, some preliminaries: Recall that the diagonal map $\Delta\co X\to X\times X$ is defined as $x\stackrel{\Delta}{\longmapsto}(x,x)$ Also recall that in a topology generated by a basis (like the product topology), a set $Y$ is open if and only if, for every point $y\in Y$ there's a basis element $B$ with $y\in B\subset Y$ Basis elements for $X\times X$ have the form $U\times V$ where $U,V$ are open sets in $X$

Now, suppose that $X$ is Hausdorff. We'd like to show its image under $\Delta$ is closed. We can do that by showing that its complement $\Delta(X)^c$ is open. $\Delta(X)$ consists of points with equal coordinates, so $\Delta(X)^c$ consists of points $(x,y)$ with $x$ and $y$ distinct.

For any $(x,y)\in \Delta(X)^c$ the Hausdorff condition gives us disjoint open $U,V\subset X$ with $x\in U, y\in V$ Then $U\times V$ is a basis element containing $(x,y)$ $U$ and $V$ have no points in common, so $U\times V$ contains nothing in the image of the diagonal map: $U\times V$ is contained in $\Delta(X)^c$ So $\Delta(X)^c$ is open, making $\Delta(X)$ closed.

Now let's suppose $\Delta(X)$ is closed. Then $\Delta(X)^c$ is open. Given any $(x,y)\in\Delta(X)^c$ there's a basis element $U\times V$ with $(x,y)\in U\times V\subset\Delta(X)^c$ $U\times V$ lying in $\Delta(X)^c$ implies that $U$ and $V$ are disjoint.

If we have $x\neq y$ in $X$ then $(x,y)$ is in $\Delta(X)^c$ The basis element containing $(x,y)$ gives us open, disjoint $U,V$ with $x\in U, y\in V$ $X$ is Hausdorff, just like we wanted.