Regular implies preregular

Statement
Any regular space is a preregular space.

Proof
Given: A topological space $$X$$ that is regular. Topologically distinguishable points $$x,y \in X$$.

To prove: There exist disjoint open subsets $$U \ni x, V \ni y$$.

Proof: Because $$x,y$$ are topologically distinguishable, it is true that either $$x \notin \overline{ \{ y \} }$$ or $$y \notin \overline{ \{ x \} }$$ (both may also be true). it suffices to consider only the first case, because the result we want to prove is symmetric in $$x$$ and $$y$$.

In the first case, let $$A = \overline{ \{ y \} }$$. Now use the definition of regularity on $$x$$ and $$A$$ to obtain the open subsets $$U$$ and $$V$$ satisfying the desired conditions.