Let $X$ be a $T_1$-space, and show that $X$ is normal if and only if each neighbourhood of a closed set $F$ contains the closure of some neighbourhood of $F$.
I saw this statement in the book Introduction to topology and modern analysis by Simmons. I tried to prove it but I did not get any where. How may it be proved?
So you want to show the following: $X$ normal is equivalent to
(*) for all closed subsets $F$ of $X$ and every open set $U$ with $F \subseteq U$ there exist some open set $V$ such that $F \subseteq V \subseteq \overline{V} \subseteq U$.
To show(*) from normality suppose we have $F \subseteq V$ as in the premise.
Then $F$ and $G := X\setminus V$ are closed and disjoint. So by normality we have open subsets $O_1$ and $O_2$ that are disjoint and such that $F \subseteq O_1$ and $X \setminus V \subseteq O_2$ The latter means that $X \setminus O_2 \subseteq V$ (And the former set is closed) and also $O_1 \subseteq X\setminus O_2$ by disjointness, we get
$$F \subseteq O_1 \subseteq \overline{O_1} \subseteq \overline{X \setminus O_2} = X \setminus O_2 \subseteq V\text{,}$$
so we can take $U = O_1$.
On the other hand if ($\ast$) holds and $F, G$ are closed and disjoint, apply (*) to $F$ and $U = X \setminus G$, and take $O_1 = V$ and $O_2 = X\setminus \overline{V}$ to get disjoint open neighbourhoods of $F$ and $G$.