Every uniformizable space is regular

Since $X\setminus C$ is an open neighbourhood of $x$, choose an entourage $E$ with $E[x]\subseteq X\setminus C$, then choose a symmetric $D$ with $D^{\circ2}\subseteq E$.

Let $O$ be the union of all open subsets of $D[x]$ containing $x$. Then $O$ is open and $x\in O\subseteq D[x]$ because $D[x]$ is a neighbourhood.

One has $\overline O\subseteq E[x]$. Indeed, if $y\notin E[x]$, then the neighbourhood $D[y]$ is disjoint from $O$: a point $z\in D[y]\cap O\subseteq D[y]\cap D[x]$ would give $(x,y)\in D^{\circ2}\subseteq E$ by symmetry. Hence $y\notin\overline O$ by [L3].

Since $E[x]\cap C=\varnothing$, step 2.1 gives $C\subseteq X\setminus\overline O$. The two open sets $O$ and $X\setminus\overline O$ are disjoint neighbourhoods of $x$ and $C$, so the space is regular by [L2].