Proof concerning regular space: there exist a closed set contained in any open set containing $x$ I was given a claim:

Let $(X, \mathfrak{T})$ be a topological space. Then $X$ is a regular
   space iff $\forall x \in X, \forall U \in \mathfrak{T}$ s.t. $x \in
 U$, $\exists V$ such that $x \in V \subseteq \overline V \subseteq U$

Regular is defined as (similar to wiki):

If, given any point $x$ and closed set $F$ in $X$ such that $x$ does
  not belong to $F$, they are separated by open disjoint neighbourhoods

I was stuck on a minor part of the proof and need some help!

$(\Rightarrow)$
Let $(X, \mathfrak{T})$ be a regular space. Then $\forall x \in X$, $F \subset X$, $F$ is closed and does not contain $x$, there exists open sets $U$ and $W$, such that $x \in U, F \subset W$ and $U \cap W = \varnothing$.
Since $U$ is open, for all points $X$, there exists an open set $V$ such that $x \in V \subset U$. (We have just produced $V$ in the claim)
Let $y \in F \subset W$ be another point. Then by definition of regular, $y$ is separated from any closed set. Consider the closed set $\overline V$, then we have $y \in W, x \in V \subseteq \overline V \subseteq U$. And since $U \cap W = \varnothing, y \notin \overline V$
($\qed$ $\eop$ <--- how to generate end of proof symbol?)


There is something terribly wrong in the second last line of the proof
  which I stated:
" $x \in V \subseteq \overline V \subseteq U$"
There is no insurance that such $\overline V \subseteq U$. When is it true that $V \subseteq \overline V \subseteq U$ will hold for open sets? It is very likely that $U \subseteq \overline V$ instead.
How do I correct this line?

 A: Here's an outline for a proof of your claim assuming $X$ is regular.
Let $x\in X$ be any point and let $U$ be some open set containing $x$. We know that $X\setminus U$ is closed so by regularity there exist disjoint open sets $V,W$ with $x\in V$ and $X\setminus U\subseteq W$. Let $y$ be a limit point of $V$. The point $y$ is not in $W$, because otherwise we would find an open set $U'$ such that $y\in U'\subseteq W$, but this contradicts that $V\cap 
W=\emptyset$ since $U'$ must contain points of $V$. Hence $y\not\in W$ which implies $y\in U$ and it follows $\overline{V}\subseteq U$.
A: You seem to want to prove that, for a topological space $X$, the following conditions are equivalent:


*

*for all $x\in X$ and all open subsets $U$ such that $x\in U$, there exists an open set $V$ such that $x\in V$ and $V\subseteq \bar{V}\subseteq U$

*for all $x\in X$ and all closed subsets $F$ such that $x\notin F$, there are disjoint open sets $U$ and $V$ such that $x\in V$, $F\subseteq U$
Proof of 1$\implies$2
Let $x\in X$ and let $F$ be a closed subset with $x\notin F$. Then $W=X\setminus F$ is an open set and $x\in W$; by hypothesis, there exists an open set $V$ with $x\in V$ and $V\subseteq\bar{V}\subseteq W$. Then $U=X\setminus\bar{V}$ is an open set and $V\cap U=\emptyset$.
Proof of 2$\implies$1
Let $x\in X$ and let $U$ be an open set with $x\in U$. Then $F=X\setminus U$ is a closed set and $x\notin F$. By assumption, there are disjoint open sets $V$ and $W$ such that $x\in V$ and $F\subseteq W$. The second condition means $U=X\setminus F\supseteq X\setminus W=G$. Since $V\cap W=\emptyset$, we have $V\subseteq G$; therefore $\bar{V}\subseteq G$. Thus $V\subseteq\bar{V}\subseteq U$.

You're not choosing carefully the sets, this is the source of your problems.
