An equivalent condition for perfectly normal topological space A topological space $X$ is
called a perfectly normal space if $X$ is a normal space
and every closed subset of $X$ is a $G_\delta$-set, equivalently
a normal space X is perfectly normal
 if every open subset of $X$ is an $F_\sigma$-set.
  The following is an exercise in page 49 of Ryszard Engelking's General topology that I cannot prove the if
  part

Show that a $T_1$-space $X$ is perfectly normal if and only if
  for every open set $W \subseteq X$ there exists a sequence $W_1,
W_2,...$ of open subsets of $X$ such that $W =\cup_{i=1}^\infty
W_i$ and $\overline{W_i}\subseteq W$ for $i= 1,2,...$, where
  $\overline{W_i}$ is the closure of $W_i$ in $X$.

 A: Theorem 1.5.15 in Engelking proves:
Let $X$ be a $T_1$ space such that for every closed set $F$ and every open set $W$ with $F \subseteq W$, there exists a sequence $W_1, W_2, \ldots$ of open sets such that $F \subseteq \cup_i W_i$ and $\overline{W_i} \subseteq W$, then $X$ is normal.
(where normal is $T_4$ in Engelking hence the $T_1$ which is included).
Then your condition already implies that for open $W$, $W = \cup \overline{W_i}$ (every $x \in W$ is in some $W_i \subseteq \overline{W_i}$, and all $\overline{W_i}$ are subsets of $W$). so every open set is an $F_\sigma$, as required. And the above normality criterion is also obeyed trivially (apply your condition to the given $W$). So such $X$ has all closed sets being $G_\delta$ and is normal by this criterion. This shows sufficiency. 
Necessity: suppose $X$ is perfectly normal and let $W$ be open. Then write $O = \cup_i F_i$ as an $F_\sigma$ and apply normality to $F_i$ and $W$ to find open $W_i$ with $F_i \subseteq W_i \subseteq \overline{W_i} \subseteq W$, and it's easily shown that these are as required.
