This problem comes from Question on Normal Topological Spaces.
I was wondering if this can be done by contradiction.
Problem. Suppose that $X$ is a normal topological space, that $F\subseteq X$ is closed, and that $F\subseteq U_1 \cup U_2$ for open sets $U_1,U_2$. Prove that there exist closed sets $F_1,F_2$ with $F=F_1 \cup F_2$, $F_1 \subseteq U_1$, and $F_2 \subseteq U_2.$
Attempt.
Since X is normal, then for every $F_1,F_2$ closed with $F_1\cap F_2=\varnothing$, there exists $U_1,U_2$ open, so that $F_1\subset U_1, F_2\subset U_2 $ and $U_1\cap U_2=\varnothing.$
One needs to prove that for all $F_1,F_2$ closed, $F\neq F_1 \cup F_2$, $F_1 \not\subseteq U_1$, or $F_2 \not \subseteq U_2...[1]$
One idea, I think, but I'm not sure if it's correct:
Let $F_1,F_2$ be closed.
Let $x\in F.$ Since $F\neq F_1 \cup F_2,$ then 1) $F\not\subset (F_1 \cup F_2)$ or 2) $ (F_1 \cup F_2)\not\subset F.$
Suppose 1), then $F_1\cap F_2=\varnothing.$ Thus one can apply normality i.e. there exists $U_1,U_2$ open, so that $F_1\subset U_1, F_2\subset U_2 $ and $U_1\cap U_2=\varnothing.$
But with this, there'd be a contradiction e.g. with one assumption $F_1 \not\subseteq U_1$...(1)
Is that correct?
A no no, (1) is not correct because there's an or in [1], so at least 2 of them could indeed be false.. sorry, I just noticed about this.
Consider the closed sets $F\setminus U_1$ and $F\setminus U_2$. Check that these are disjoint. By normality there exest open sets $V_1,V_2$ such that $V_1\cap V_2=\emptyset$ and $F\setminus U_1 \subset V_1$,$F\setminus U_2 \subset V_2$. Now show that $F_1=F\setminus V_1$ and $F_2=F\setminus V_2$ satsifies your requirements.