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).