topology space $T_4$ question Prove that A topology space $X$ is $T_4$ if and only if for all disjoint closed subsets $F$ and $F$ of $X$ there exist open subset $U$ and $V$ of $X$ such that
$F \subset U$ , $F' \subset V$, and $CL(U) \cap CL(V)=\emptyset$
I can do the backward direction but the forward Im not so sure. Here is what is got.
Assume that $X$ is $T_4$, this means for all disjoint closed subsets $F$ and $F$ of $X$ there exist open subset $U$ and $V$ of $X$ such that
$F \subset U$ , $F' \subset V$, and $U \cap V=\emptyset$
How can I used this to prove that $CL(U) \cap CL(V)=\emptyset$
 A: Essentially, rinse and repeat.
Recall that another equivalent formalution of normality is that if $\DeclareMathOperator{\clos}{CL}F \subseteq U \subseteq X$ are such that $F$ is closed and $U$ is open, then there is an open $V \subseteq X$ with $F \subseteq V \subseteq \clos (V) \subseteq U$.
So we'll make three applications of normality:


*

*Since $F , F^\prime$ are disjoint closed sets, by normality there are open sets $W$ and $W^\prime$ such that $F \subseteq W$, $F^\prime \subseteq W^\prime$ and $W \cap W^\prime = \varnothing$.

*Since $F \subseteq W$ with $F$ closed and $W$ open, by the above characterisation of normality there is an open set $G$ such that $F \subseteq G \subseteq \clos (G) \subseteq W$.

*Since $F^\prime \subseteq W^\prime$ with $F^\prime$ closed and $W^\prime$ open, by the above characterisation of normality there is an open set $G^\prime$ such that $F^\prime \subseteq G^\prime \subseteq \clos (G^\prime) \subseteq W^\prime$.
Now among the four open sets listed above, find two to label as $U$ and $V$ so that 


*

*$F \subseteq U$; 

*$F^\prime \subseteq V$; and 

*$\clos (U) \cap \clos (V) = \varnothing$.

