Combine two ways of formulating normality:
$X$ is normal (in the sense of separating two disjoint closed sets by disjoint open sets) iff for all closed sets $F$ and open sets $U$ such that $F \subseteq U$ we have an open subset $V$ such that $F \subseteq V \subseteq \overline{V} \subseteq U$.
Proof: This is quite easy to see: suppose $F$ and $U$ with $F \subseteq U$ are given, then $F$ and $U^\complement$ are disjoint and closed so we have disjoint open sets $O_1$ and $O_2$ such that $F \subseteq O_1$ and $U^\complement \subseteq O_2$. But then $O_1 \subseteq O_2^\complement$ by disjointness of $O_1$ and $O_2$ and also $O_2^\complement \subseteq U$; as $O_2^ \complement$ is closed and contains $O_1$, $$F \subseteq O_1 \subseteq \overline{O_1} \subseteq O_2^ \complement \subseteq U$$ and we can take $V=O_1$.
For the reverse, of we have this property and two disjoint closed sets $F_1$ and $F_2$, we have $F_2^ \complement$ open and containing $F_1$ so we have $V$ open with $$F_1 \subseteq V \subseteq \overline{V} \subseteq F_2^\complement$$ and then $O_1= V$ and $O_2=\overline{V}^ \complement$ are the required disjoint open sets separating $F_1$ and $F_2$. QED.
Knowing this, it's easy: separate $F_1$ and $F_2$ by disjoint open sets $O_1$ and $O_2$ resp. and then apply the alternative formulations again for both $F_i$ and $O_i$ ($i=1,2$) to get open neighbourhoods $U_i$ of $F_i$ with closure inside the $O_i$, so a fortiori disjoint too.
