The topology of a uniformizable space is always a symmetric topology; that is, the space is an $R_{0}$-space.
How to prove it? It must be a simple question, but I can't write it down.
$X$ is an $R_{0}$ space if any two topologically distinguishable points in $X$ are separated.
Two points of a topological space $X$ are topologically indistinguishable if they have exactly the same neighborhoods.
Two points of $X$ are topologically distinguishable if they are not topologically indistinguishable, this means there is an open set containing precisely one of the two points.
Thanks a lot.
HINT: Let $\mathscr{U}$ be a diagonal uniformity on $X$. As usual, for $U\in\mathscr{U}$ and $x\in X$ let
$$U[x]=\{y\in X:\langle x,y\rangle\in U\}\;;$$
then the topology on $X$ induced by $\mathscr{U}$ is the set of $V\subseteq X$ such that for each $x\in V$ there is a $U_x\in\mathscr{U}$ such that $x\in U_x[x]\subseteq V$. Now use the fact that $\mathscr{U}$ is symmetric: $U\in\mathscr{U}$ if and only if $U^{-1}=\{\langle y,x\rangle\in X\times X:\langle x,y\rangle\in U\}\in\mathscr{U}$.