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 $V\cap V'=\emptyset$?