Munkres Lemma 31.1, no need for $T_1$ 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.

 A: 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.
A: Munkres' definition of regularity requires that the space be $T_1$. Many authors do not include $T_1$ in the definition and call a space $T_3$ if it is regular (in the "weaker" sense)  and $T_1$. 
See https://en.wikipedia.org/wiki/Separation_axiom
Anyway, to prove the core of the lemma, you do not need $T_1$. The same is true for normal spaces. It is again a matter of definitions.
