A topological space $(X, \mathscr{T})$ is uniformizable if there exists a uniform structure on $X$ that induces $\mathscr{T}$.
I am trying to prove that every uniformizable space on $X$ is T3. To do this, I have to show that:
For every $p \in X$ and $Q \subseteq X$ such that $Q$ is closed and that $p \not \in Q$, there exist open sets $U, V \subseteq X$ such that $$p \in U \text{ and } Q \subseteq V \text{ and } U \cap V = \varnothing.$$
I'm doing the following: for any $p$ and $Q$ that satisfy the hypothesis, it follows that $Q^c$ (the complement of $Q$) is an open set containing $p$. Therefore, there exists an entourage $E$ in the uniform structure such that $Ep \subseteq Q^c$, where $Ep$ is the set of $E$-left-relatives of point $p$, i.e. $$Ep = \{q \in X \mid (q, p) \in E \}.$$ From here, however, I'm not entirely sure how to find an open superset of $Q$ that is disjoint with $Ep$.
I'm looking for a small hint, perhaps pointing me to the proper axiom to use to construct the applicable open sets. Any help is appreciated.
To show that a space is a $T_3$ space (regular for those using the other nomenclature), we need to show that the closed neighbourhoods form a neighbourhood basis of each point.
In that form, it is easier to show, I think.
So let $x \in X$, and $W$ an arbitrary entourage. We must show that
$$W(x) = \{ y : (x,y) \in W\}$$
contains a closed neighbourhood of $x$. Let $V$ be a symmetric entourage with $V^3 \subset W$ ...