It should be a triviality, I believe.
The topology induced by a uniform structure $\mathcal{U}$ with $\cap \mathcal{U} =\Delta$, where $\Delta$ is the diagonal, is Hausdorff.
Now I think that if I define the product topology $X \times X$ to be such that the closed sets of it are the sets that are in the $\mathcal{U}$, then I will get that $\Delta$ is closed as an arbitrary intersection of closed sets, and thus the space $X$ is $T_2$.
Is this right? (I am just not sure that if I induce the sets in $\mathcal{U}$ as open sets in $X \times X$ that the above will occur).
You don’t need to look at any topology on $X\times X$; just show Hausdorffness of $X$ directly. Let $x$ and $y$ be distinct points of $X$. Then $\langle x,y\rangle\notin\Delta$, so there is some $U\in\mathcal{U}$ such that $\langle x,y\rangle\notin U$. Now use the uniformity axioms to show that there is a symmetric $V\in\mathcal{U}$ such that $V\circ V\subseteq U$, and then show that $V[x]$ and $V[y]$ must be disjoint nbhds of $x$ and $y$, respectively.