A space is regular if and only if every point has a neighbourhood base of closed neighbourhoods, if and only if $x \in U$ open gives an open $V$ with $x \in V \subseteq \overline{V} \subseteq U$

Assume (a) and let $U$ be open with $x \in U$; then $C := X \setminus U$ is closed by [L4] and $x \notin C$, so [A1] gives disjoint open $V \ni x$ and $W \supseteq C$.

Assume (b) and let $N$ be a neighbourhood of $x$; fix an open $U$ with $x \in U \subseteq N$ by [L1], and let $V$ be as in (b), so $x \in V \subseteq \overline{V} \subseteq U \subseteq N$.

Assume (c) and let $C$ be closed with $x \notin C$; then $X \setminus C$ is open by [L4] and contains $x$, hence is a neighbourhood of $x$ by [L1], so (c) gives a closed neighbourhood $K$ of $x$ with $K \subseteq X \setminus C$.

Under step 1.1: $V \subseteq X \setminus W$, since $V$ and $W$ are disjoint, and $X \setminus W$ is closed by [L4], so $\overline{V} \subseteq X \setminus W$ by [L2]; and $X \setminus W \subseteq X \setminus C = U$ because $C \subseteq W$.

Under step 1.2: $\overline{V}$ is a closed set containing the open $V \ni x$, so it is a neighbourhood of $x$ by [L1], and it is a closed neighbourhood of $x$ contained in $N$.

Under step 1.3: put $V_0 := \operatorname{int}(K)$, which is open and contains $x$ by [L3] since $K$ is a neighbourhood of $x$; and put $W_0 := X \setminus K$, which is open by [L4] since $K$ is closed.

Step 2.1 gives $x \in V \subseteq \overline{V} \subseteq U$ with $V$ open, so (a) implies (b).

Step 2.2 gives, for every neighbourhood $N$ of $x$, a closed neighbourhood of $x$ inside $N$, so (b) implies (c).

Under step 2.3: $V_0 \cap W_0 = \operatorname{int}(K) \cap (X \setminus K) = \varnothing$ because $\operatorname{int}(K) \subseteq K$ by [L3], and $C \subseteq X \setminus K = W_0$ because $K \subseteq X \setminus C$; so $V_0$ and $W_0$ are disjoint open sets containing $x$ and $C$ respectively, and (c) implies (a).

By steps 3.1, 3.2 and 3.3 the three conditions (a), (b) and (c) are equivalent.