In Munkres we have the following lemma:
Lemma 31.1. Let $X$ be a topological space. Let one-point sets in $X$ be closed.
(a) $X$ is regular ii and only if given a point $x$ of $X$ and a neighborhood $U$ of $x$,
there is a neighborhood $V$ of $x$ such that $\operatorname{cl}(V)\subseteq U$.
I do not see the use of $T_1$ assumption (Let one-point sets in $X$ be closed.) in the proof. I proved it myself before finding the result in Munkres without needing $T_1$.
(1) Could you confirm my statement?
(2) Can we say the same for the case of normal spaces (Lemma 31.1(b))? I've yet to start with the proof.
Best Answer
Let $S = \{0,1\}$ be the Sierpinski space with $0$ being the open point. This satisfies the condition on neighbourhoods imposed by the lemma, and so we should have that $S$ is regular, which is not true if your definition of regular assumes $T_1$.
Otherwise, I think this should work, with the following proof:
Suppose that $X$ is regular and fix an open set $U \ni x$, so that there exist disjoint open sets $V \ni x, W \supset U^c$. This imples $V\subset W^c$, and so $cl(V) = cl(W^c) = W^c \subset U$.
Reciprocally, take $x \not \in F$ with $F$ closed. By hypothesis there exists $x \in V \subset F^c$ open such that $cl(V) \subset F^c$, and this implies that $cl(V)^c \supset F$. Now take $V$ and $cl(V)^c$ as your separating neighbourhoods.