Prove a topological space is normal if and only if for each pair of open sets $U,V$ such that $U cup V=X$, there exists open sets $U’$ and $V’$ such that $overline U’ subset U$ and $overline V’ subset V$ and $U’ cup V’=X$

Attempt:

$(Rightarrow)$ Since $U cup V=X implies (X-U) cap (X-V)=varnothing$. $X-U$ and $X-V$ are disjoint closed sets. Then by normality of $X$, there are disjoint open sets $W,Z$ with $X-U subset W$ and $X-V subset Z$, and an equivalent definition of normality states there are open sets $W’,Z’$ containing $X-U$ and $X-V$ with $overline W’ subset W$ and $overline Z’ subset Z$. Now I am unsure how to show that $W’ cup Z’=X$, which is what I need to show in the problem.

$(Leftarrow)$ Suppose the second part of the statement holds. Let $A$ be closed in $X$ and $U$ be any open set containing $X$. Then the existence of an open set $U’$ containing $A$ such that $overline U’ subset U$ implies $X$ is normal.

Comment:I am having difficulty understanding (why/if) $W’ cup Z’=X$ holds in the first direction. Also I am skeptical whether I started the second direction correctly by starting with a closed set, and an arbitrary open set containing this closed set. I also realize it may not be true that $U’$ contains the closed set $A$, which adds to my skepticism. Any ideas?