Hausdorffness is closure-local

Statement
Suppose $$X$$ is a topological space such that for any point $$x \in X$$, there exists an open subset $$U \ni x$$ such that the closure $$\overline{U}$$ is a Hausdorff space. Then, $$X$$ is also a Hausdorff space.

Related facts

 * Hausdorffness is not local
 * Hausdorffness is hereditary

Facts used

 * 1) uses::Openness is transitive

Proof
Given: A topological space $$X$$ such that for any point $$x \in X$$, there exists an open subset $$U \ni x$$ such that the closure $$\overline{U}$$ is a Hausdorff space. We are given two distinct points $$x,y \in X$$.

To prove: We can find disjoint open subsets $$V,W$$ of $$X$$ such that $$V \ni x, W \ni y$$.

Proof: