# $X$ is normal if and only if $A\subseteq U$ implies there exists $V$ such that $A\subseteq V\subseteq\overline{V}\subseteq U$

I want to prove this: $X$ is normal if and only if for every closed $A$ and open $U$ such that $A\subseteq U$, there exists $V$ such that $A\subseteq V\subseteq\overline{V}\subseteq U$.

Suppose $X$ is normal. Let $A\subseteq U$ where $A$ is closed and $U$ is open. Then $X\setminus U$ is closed and $(X\setminus U)\cap A=\emptyset$. But since $X$ is normal, there exist open sets $V$ and $V'$ such that $A\subseteq V$, $X\setminus U\subseteq V'$, and $V\cap V'=\emptyset$. Clearly $V\subseteq \overline{V}$, and also $V\subseteq U$ because $V\cap V'=\emptyset$.

But how does that guarantee the closure of $V$ is also contained in $U$?

Conversely, suppose for each $A\subseteq U$ with $A$ closed and $U$ open, there exists an open set $V$ such that $A\subseteq V\subseteq\overline{V}\subseteq U$. Let $C_1, C_2$ be closed and $C_1\cap C_2=\emptyset$. Then $X\setminus C_2$ is open and $C_1\subseteq X\setminus C_2$. By assumption, there exists an open $V$ such that $C_1\subseteq V\subseteq\overline{V}\subseteq U$, hence $V\cap C_2=\emptyset$. Similarly there exists open $V'$ such that $C_2\subseteq V'\subseteq\overline{V'}\subseteq U$ with $V'\cap C_1=\emptyset$.

But how do we get that $V\cap V'=\emptyset$?

For the forward direction, we let $$A\subseteq U$$ for some closed $$A$$ and open $$U$$. Then, $$U^c$$ is closed, and $$A\cap U^c = \emptyset$$, as you pointed out. Therefore, we have some disjoint open $$V_1$$ and $$V_2$$ such that $$A\subseteq V_1$$ and $$U^c\subseteq V_2$$. As orangeskid points out, $$V_1\cap V_2 = \emptyset$$ implies $$V_1\subseteq V_2^c\subseteq U$$, and $$V_2^c$$ is closed, so $$\overline{V_1}\subseteq V_2^c\subseteq U$$. Therefore, $$A\subseteq V_1\subseteq \overline{V_1}\subseteq U$$.
You might also consider an application of Urysohn's lemma: let $$f : X\to [0, 1]$$ be a continuous function such that $$f(A) = \{1\}$$ and $$f(U^c) = \{0\}$$. Then, consider that $$S = f^{-1}((1/2, 1])$$ (or $$S = f^{-1}((a, 1])$$ for any $$0 < a < 1$$) is open in $$X$$ and satisfies $$A\subseteq S\subseteq U$$. The fact that $$\overline{S}\subseteq U$$ follows from $$f(\overline{S})\subseteq [1/2, 1]$$ (which implies that $$\overline{S}\cap U^c = \emptyset$$).
For the backward direction, we let $$C_1$$ and $$C_2$$ be two disjoint closed subsets of $$X$$ as you did. Then, we consider some open $$V$$ such that $$C_1\subseteq V\subseteq \overline{V}\subseteq C_2^c$$. As $$(\overline{V})^c$$ is open, is disjoint from $$V$$, and contains $$C_2$$, we have disjoint open neighborhoods $$V\supseteq C_1$$ and $$(\overline{V})^c\supseteq C_2$$, and therefore $$X$$ is normal.
Since $V\cap V'= \emptyset$, we have $V \subset X \backslash V'$, a closed set, so $\bar V \subset X\backslash V'$.