Show that $(X,\tau_X)$ is Hausdorff if and only if the diagonal $\{(x,x) : x \in X\}$ is closed in $(X\times X,\tau_{X \times X})$.

Question

I already proved that if $(X,\tau_X)$ is Hausdorff then $(X\times X,\tau_{X \times X})$ is husdorff. Now what I did is that, since we know singletons are closed in Hausdorff spaces, I denote $U_{(x,x)}=X\times X\setminus \{x,x\}$. This is an open set by the above property. Now I want to show that $\bigcap_{x\in X}U_{(x,x)}=\bigcup_{x\in X}X\times X\setminus \{x,x\}=diagonal$ is closed. But I can't figure out how to prove that the intersection is open since the arbitrary intersection of open sets are not necessarily open. Any help would be appreciated, thank you!