Every $T_3$ space with a countable basis is $T_4$.

Question

Theorem. Every regular space with a countable basis is normal.

Proof. Let X be a regular space with a countable basis $\mathcal{B}$. Let $A$ and $B$ be disjoint closed subsets of $X$. Each point $x$ of $A$ has a neighborhood $U$ disjoint from $B$. Using regularity, choose a neighborhood $V$ of $x$ whose closure lies in $U \dots$

I can't follow the last step made until this point; Using regularity, choose a neighborhood $V$ of $x$ whose closure lies in $U$.

I do know that $X$ is a regular space if, given any nonempty closed set $F$ and any point $x$ that does not belong to $F$, there exists a neighbourhood $U$ of $x$ and a neighbourhood $V$ of $F$ that are disjoint.

Could anyone help me with some insights?

Thanks in advance.

Answer 1

$x$ 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.