Suppose that $X$ is a uniform space with uniformity $U$. We can make a topology as follows: for any $x\in X$ the set $\{u[x]:u\in U\}$ is a nhood base at $x$. Here $u[x] = \{y\in X: (x,y)\in u\}$.
Is the generated topology always $T_1$? I know that the generated topology is $T_2$ if and only if $\bigcap\limits_{u\in U} u = \{(x,x): x\in X\}$. Is it always $T_1$, or can we not say this?
We have the trivial uniform structure $\mathscr{U} = \{X\times X\}$, which induces the trivial topology. Unless $\operatorname{card} X \leqslant 1$, this is not a $T_1$ topology.
Apart from that, every topology induced by a uniform structure is a $T_{3\frac{1}{2}}$ topology (or completely regular, depending on the used nomenclature), and a $T_{3\frac{1}{2}}$-space is Hausdorff if and only if it is $T_0$, so for topologies induced by uniform structures, $T_0,\, T_1,$ and $T_2$ are equivalent.