If $(X,\tau)$ is a normal space then any closed subspace of $(X,\tau)$ is normal.
Let $(X,\tau)$ is normal and $A\subseteq X$ is closed in $(X,\tau)$. We must show that $(A,\tau_A)$ is normal.
Let $B \subseteq A$ $\subseteq X$ and $C \subseteq A$ $\subseteq X$ .Take $\overline B$ (closure of B) and $\overline C$ from $(X,\tau)$ such that $\overline B$ $\cap $ $\overline C$=$\emptyset$ . (They are always closed)
Since A is closed $\overline B$ $\cap$ $ A $ and $\overline C$ $\cap $ $A $ are closed in $(X,\tau)$. Since they are closed there exists disjoint subsets $O_1$ and $O_2$ such that $\overline B$ $\cap$ $ A $ $\subseteq $$O_1$ and $\overline C$ $\cap $ $A $ $\subseteq $ $O_2$ because $(X,\tau)$ is normal space.
$\overline B^A$ = $\overline B$ $\cap$ $ A $ , $\overline C^A$ = $\overline C$ $\cap $ $A$ we have $\overline B^A$ $\subseteq $ $O_1$ and $\overline C^A$ $\subseteq $ $O_2$
If we take intersection with $A$ from both sides of $\subseteq $ in $\overline B^A$ $\subseteq $ $O_1$ and $\overline C^A$ $\subseteq $ $O_2$ we have $\overline B^A$ $\subseteq $ $O_1 \cap A$ and $\overline C^A$ $\subseteq $ $O_2 \cap A$
Since $\overline B^A$ (closure of B in $(A,\tau_A)$) and $\overline C^A$ are closed disjoint sets in $(A,\tau_A)$ and $O_1 \cap A$ and $O_2 \cap A$ disjoint open sets in $(A,\tau_A)$ we have $(A,\tau_A)$ is normal too.
What is the mistake in this proof or what are the missings? Could someone correct me please? Thanks in advance.
Best Answer
You overcomplicate things:
If $A \subseteq X$ is closed and $C,D \subseteq A$ are closed (and disjoint) in $A$, then they're also closed in $X$.
So we separate them by open sets $U$ of $V$ of $X$ as $X$ is normal, and then $U \cap A$ and $V \cap A$ are the separating open sets in $A$.