Here is Prob. 2, Sec. 31, in the book Topology by James R. Munkres, 2nd edition:
Show that if $X$ is normal, every pair of disjoint closed sets have neighborhoods whose closures are disjoint.
Here is the definition of normal space:
A topological space $X$ is said to be normal if
(i) singleton subsets (and hence finite subsets) of $X$ are closed in $X$, and
(ii) for every pair $A$ and $B$ of disjoint closed sets in $X$, there are disjoint open sets $U$ and $V$ in $X$ containing $A$ and $B$, respectively.
My Attempt:
Let $A$ and $B$ be any two disjoint closed sets in a normal space $X$. Then there exist disjoint open sets $U$ and $V$ of $X$ such that $A \subset U$ and $B \subset V$.
Now as $A$ is a closed set in the normal space $X$ and as $U$ is an open set containing $A$, so by Lemma 31.1 (b) in Munkres there is an open set $U^\prime$ in $X$ such that $A \subset U^\prime$ and $\overline{U^\prime} \subset U$. Thus we have
$$ A \subset U^\prime \subset \overline{U^\prime} \subset U. $$Similarly, as $B$ is a closed set in the normal space $X$ and as $V$ is an open set containing $B$, so there exists an open set $V^\prime$ in $X$ such that
$$ B \subset V^\prime \subset \overline{V^\prime} \subset V. $$Finally, as $U$ and $V$ are disjoint and as $\overline{U^\prime} \subset U$ and $\overline{V^\prime} \subset V$, so the sets $\overline{U^\prime}$ and $\overline{V^\prime}$ are also disjoint.
Thus $U^\prime$ and $V^\prime$ are neighborhoods of sets $A$ and $B$ (i.e. open sets containing these sets), respectively, such that their closures $\overline{U^\prime}$ and $\overline{V^\prime}$ are disjoint.
The above proof is very similar to this one.
Is this proof correct? If so, is it clearly enough presented? If not, then where is it deficient?
Last but not least, does the converse also hold?
Best Answer
You ask also whether the converse statement:
A topological space $X$ such that every two disjoint closed sets have neighbourhoods with disjoint closure is normal.
This is clearly true: if $A, B \subset X$ are closed disjoint sets, by assumption there are open sets $A \subset U$, $B \subset V$ such that $\overline{U} \cap \overline{V} = \emptyset$. Then $U \cap V = \emptyset$ and $X$ is normal.