Definition 4.9.1 (Regular Space, [Proposition 1.4.11, Bou13]). Let $X$ be a topological space, then the following are equivalent:
For each $x \in X$ and $A \subset X$ closed with $x \not\in A$, there exists $U \in \cn(x)$ and $V \in \cn(A)$ such that $U \cap V = \emptyset$.
For each $x \in X$, the closed neighbourhoods of $x$ forms a fundamental system of neighbourhoods at $x$.
If $X$ is a T1 space such that the above holds, then $X$ is regular.
Proof. $(1) \Rightarrow (2)$: Let $U \in \cn^{o}(x)$, then $U^{c}$ is closed with $x \not\in U^{c}$ by (V1). Thus there exists $V \in \cn^{o}(U^{c})$ such that $x \not\in V$, so $V^{c} \in \cn(x)$ is closed with $V^{c} \subset U$.
$(2) \Rightarrow (1)$: Since $X$ is Hausdorff, $X$ is T1 as well. Let $x \in X$ and $A \subset X$ closed such that $x \not\in A$, then $A^{c} \in \cn(x)$. So there exists $K \in \cn(x)$ closed such that $x \in K \subset A^{c}$. Thus $K \in \cn(x)$ and $K^{c} \in \cn(A)$ are the desired neighbourhoods.$\square$