If $X$ is normal and $A$ is a $F_{\sigma}$-set in $X$, then $A$ is normal. How could I prove this theorem? A topological space $X$ is a normal space if, given any disjoint closed sets $E$ and $F$, there are open neighbourhoods $U$ of $E$ and $V$ of $F$ that are also disjoint. (Or more intuitively, this condition says that $E$ and $F$ can be separated by neighbourhoods.)
And an $F_{\sigma}$-set is a countable union of closed sets.
So I should be able to show that the $F_{\sigma}$-set has the necessary conditions for a $T_4$ space? But how could I for instance select two disjoint closed sets from $F_{\sigma}$?

Let us begin with a lemma ( see it in the Engelking's "General Topology" book, lemma 1.5.15):

If $X$ is a $T_1$ space and for every closed $F$ and every open $W$ that contains $F$ there exists a sequence $W_1$,  $W_2$, ... of open subsets of $X$ such that $F\subset \cup_{i}W_i$ and $cl(W_i)\subset W$ for $i=$ 1, 2, ..., then the space $X$ is normal.

Suppose $X$ is normal and $A = \cup_n F_n \subset X$ is an $F_\sigma$ in $X$, where all the $F_n$ are closed subsets of $X$. Then $A$ is normal (in the subspace topology).
To apply the lemma, let $F$ be closed in $A$ and $W$ be an open superset of it (open in $A$). Let $O$ be open in $X$ such that $O \cap A = W$, and note that each $F \cap F_n$ is closed in $X$ and by normality of $X$ there are open subsets $O_n$ in $X$, for $n \in \mathbb{N}$ such that
$$ F \cap F_n \subset O_n \subset \overline{O_n} \subset O $$ and define $W_n = O_n \cap A$, which are open in $A$ and satisfy that the $W_n$ cover $F$ (as each $W_n$ covers $F_n$ and
$F = F \cap A = \cup_n (F \cap F_n)$) and the closure of $W_n$ in $A$ equals $$\overline{W_n} \cap A = \overline{O_n \cap A} \cap A \subset O \cap A = W$$
which is what is needed for the lemma.
