Union of closed normal subspaces is normal Let $X=F_{1} \cup F_{2}$ be a space such that $F_{1}$ and $F_{2}$ are closed normal spaces(as subspace to $X$)
I need to prove that $X$ is also normal
My attempt goes as follow: From normality of each $F_{i}$ I can find disjoint open sets in $X$: $U_{i},V_{i}; i=1,2$ such that $A \cap F_{i} \subset U_{i} \cap F_{i}$ 
$B \cap F_{i} \subset V_{i} \cap F_{i}$ 
then I tried to build such open spaces for A and B in X as the unions of those sets but the problem is that they might not be disjoint
I noticed that I didn't used the fact that the $F_{i}$ are closed, how do I fix this?
 A: By normality of $F_i$, you may take $U_i, V_i$ with stronger propety, namely that the closures of $U_i\cap F_i$ and $V_i\cap F_i$ in $F_i$ are disjoint. Since $F_i$ is closed in $X$, $F_i$-closure and $X$-closure of a subset of $F_i$ coincide. For instance, as $U_i$ is open neighbourhood of $A\cap F_i$, so will be $U_i\setminus \overline{V_i}$ (because $(U_i\setminus\overline{V_i})\cap F_i = (U_i\cap F_i)\setminus(\overline{V_i}\cap F_i) \supset (U_i\cap F_i)\setminus(\overline{V_i\cap F_i}) = U_i\cap F_i$).
Now you can easily construct disjoint neighbourhoods of $A$ and $B$ from this $U_i,V_i$ (and their closures).
A: From normality of each $F_i$ there exist open sets $U_i$, $V_i$, $i=1,2$ in $$ such that $A\cap F_i\subseteq U_i\cap F_i$, $B\cap F_i\subseteq V_i\cap F_i$ and $U_i\cap V_i\cap F_i=\emptyset$.
Then $U_i\cup(X\setminus F_i)$ and $V_i\cup(X\setminus F_i)$ are open sets in $X$ and $A\subseteq U_i\cup(X\setminus F_i)$, $B\subseteq V_i\cup(X\setminus F_i)$ and $(U_i\cup(X\setminus F_i))\cap(V_i\cup(X\setminus F_i))=X\setminus F_i$.
Finally take $U:=(U_1\cup(X\setminus F_1))\cap(U_2\cup(X\setminus F_2))$ and $V:=(V_1\cup(X\setminus F_1))\cap (V_2\cup(X\setminus F_2))$. Then:
$$A\subseteq U,\quad B\subseteq V\quad\text{and}\quad U\cap V=(X\setminus F_1)\cap(X\setminus F_2)=\emptyset.$$
