Let $X$ be a normal space. By definition it is a $T_1$ space and any pair of disjoint closed sets can be separated by open sets in the sense that they have disjoint neighborhoods.
What I want to prove is
$X$ is normal iff each nbd of a closed set contains the closure of some nbd of the set
To prove the first implication I took a closed set $A$ and nbd $U$ of $A$. Since $U'$ is closed and disjoint from $A$ there exist nbd $H_A$ of $A$ and nbd $H_{U'}$ of $U'$ such that they are disjoint. I feel that closure of $H_A$ is subset of $U$ but not able to prove it.
For converse part, let $A$ and $B$ be closed disjoint sets in $X$. Then for any nbd $U_A$ and $U_B$ of $A$ and $B$ respectively there exist nbd $V_A$ and $V_B$ of $A$ and $B$ such that $$ A \subseteq V_A \subseteq \bar{V_A} \subseteq U_A $$ and
$$ B \subseteq V_B \subseteq \bar{V_B} \subseteq U_B $$ If $U_A$ and $U_A$ are disjoint then we are done. If not how can I proceed further?
Please suggest any alternative approach also if possible.
In this answer a neighborhood of a set $A$ is defined to be an open set that contains $A$ as subset.
You could also define a neighborhood of a set $A$ more broadly as a subset $N$ that satisfies $A\subseteq O\subseteq N$ for some open set $O$.
But it does not matter which of the definitions is used by defining normality.
Also what is proved below can easily be modified in such a way that we arrive at a proof applying to the broader definition.
$H_A\cap H_{U^{\complement}}=\varnothing$ tells us that $H_A\subseteq(H_{U^{\complement}})^{\complement}$ and $(H_{U^{\complement}})^{\complement}$ is closed.
So we conclude that: $$\mathsf{cl}(H_A)\subseteq(H_{U^{\complement}})^{\complement}$$
Here $U^{\complement}\subseteq H_{U^{\complement}}$ or equivalently $(H_{U^{\complement}})^{\complement}\subseteq U$ so we found that:$$\mathsf{cl}(H_A)\subseteq U$$
Let $A$ be a closed set and let $U_A$ be a neighborhood of $A$.
The sets $A$ and $U_A^{\complement}$ are closed and disjoint, so disjoint open sets $U, V$ exist with $A\subseteq U$ and $U_A^{\complement}\subseteq V$.
Then: $$A\subseteq U\subseteq V^{\complement}\subseteq U_A$$
Here $U$ is a neighborhood of $A$ and this with $\mathsf{cl}(U)\subseteq V^{\complement}$ so that also:$$A\subseteq U\subseteq \mathsf{cl}(U)\subseteq U_A$$