I have some doubts about the following proof,
1. Where was used the hypothesis of countable basis?
2. To prove that $X$ is $T_4$, for all closed sets $F,G\subset $ X , there must be 2 open sets $V,V'$ such that $F\subset V,G\subset V'$ and $V\cap V'.$ In the proof given, $$V' \supset X\setminus U=F \ ?$$ and $$G=\overline V\subset U\ ?$$
Proof (made by Willie Wong MSE user at Every $T_3$ space with a countable basis is $T_4$.)
$x$ is a point. $U\ni x$ is open. $X\setminus U$ is closed. Regularity implies that there are disjoint neighborhoods $V \ni x$ and $V' \supset X\setminus U$.
We have $$ V\cap V' = \emptyset \implies V \subseteq X \setminus V' $$ Note that the right hand side is the complement of an open set and so is closed. Take the closure on both sides we have $$ \bar{V} \subseteq \overline{X\setminus V'} = X\setminus V'$$
Now since $$ X\setminus U \subseteq V'$$ from de Morgan's laws we have $$ X\setminus V' \subseteq U $$ and so $$ \bar{V} \subseteq U $$ as desired.
The proof you have copied is not a proof of the statement you refer to at all. Instead, it is a proof that if $X$ is regular and $U$ is a neighborhood of a point $x$ in $X$, then there is a neighborhood $V$ of $x$ whose closure is contained in $U$.