In a Hausdorff space, for any open set $U$, can we find another open set $V$ such that $\overline{V} \subseteq U$? 
In a Hausdorff space $X$, for any point $x \in X$ there exists a neighborhood $U$ containing $x$. Can we find another open set $V$ containing $x$ such that $\overline{V} \subseteq U$?

This is not a homework question or anything, but if it is true it would give me an easy way to prove the reverse direction of another theorem.
Loosely this reminds me of the separation axioms, but I'm not sure if this holds for Hausdorff spaces.
 A: Consider $\mathbb{R}$ with the standard topology $\tau_0$. Let $\tau:= \{V\cup (U\cap \mathbb{Q}) \mid U,V\in \tau_0\}$. It is easy to see that $\tau$ is a Hausdorff topology on $\mathbb{R}$. Denote by $X$ the set $\mathbb{R}$ together with the topology $\tau$. Pick $x\in \mathbb{Q}$ and let $U:=\mathbb{Q}$ be the open neighborhood of $x$ in $X$. Let $V$ be any $V\in \tau$ with $x\in V\subset U$. It follows from the definition of $\tau$ that there is a $W\in \tau_0$ with $V = U \cap W$. There exists an irrational number $p\in W$ since irrational numbers are dense in $\mathbb{R}$. I claim $p$ must lie in the closure $\overline{V}$ of $V$ in $X$. If not, then $O:=X\backslash \overline{V} \in \tau$ is an open neighborhood of $p$ in $X$. It follows from the construction of $\tau$ that there is an $O'\in \tau_0$ with $p\in O' \subset O$. Because rational numbers are dense in $\mathbb{R}$, we have $\emptyset \neq O'\cap W \cap \mathbb{Q} = O' \cap V$. This is a contradiction. Therefore $\overline{V}$ does not lie in $U$.
