Could you help me prove that $X$ is normal $\iff$ for all open $U, V$such that $U \cup V = X$ there are closed $F \subset U,\ G \subset V,\ F \cup G = X?$
I've been trying to prove it using other equivalent condition for $X$ to be normal, namely that $\forall F \subset U,\ \ \ $ $F$-closed, $U$-open $\exists V$-open $F \subset V \subset \overline{V} \subset U$, but I keep failing.
Could you help me with that?
Let us call a space $X$ with the property that for all open $U,\,V$ with $X = U\cup V$ there exist closed $F\subset U$ and $G\subset V$ with $X = F\cup G$ an n-space.
Every normal space is an n-space:
Let $U,\,V$ open with $X = U\cup V$. Let $A = X\setminus U$ and $B = X\setminus V$. Then $A$ and $B$ are closed and disjoint, so by normality, there are open $R \supset A$ and $S \supset B$ with $R\cap S = \varnothing$. Let $F = X \setminus R$ and $G = X\setminus S$.
Then $F$ and $G$ are closed, $F\cup G = (X\setminus R) \cup (X\setminus S) = X\setminus (R\cap S) = X$, and $F = X\setminus R \subset X\setminus A = U$, as well as $G = X\setminus S \subset X\setminus B = V$.
Every n-space is normal:
Let $A,\, B$ be disjoint closed sets in $X$. Let $U = X\setminus A$ and $V = X\setminus B$. Then $U$ and $V$ are open with $U\cup V = X \setminus (A\cap B) = X$. Since $X$ is an n-space, there are closed $F\subset U$ and $G\subset V$ with $X = F\cup G$. Let $R = X\setminus F$ and $S = X\setminus G$.
Then $R$ and $S$ are open, $R\cap S = (X\setminus F)\cap (X\setminus G) = X\setminus (F\cup G) = \varnothing$, and $A = X\setminus U \subset X\setminus F = R$, as well as $B = X\setminus V \subset X\setminus G = S$.