I've problems understanding the necessity of the $T_1$ separation axiom in a certain lemma. First I'll the definitions we are using in class. Let $X$ be a topological space.
$X$ satisfies the $T_1$ separation axiom iff every singleton is closed.
also
$X$ is regular if for each point $x$ and closed $C$ such that $x\not\in C$ there are open disjoint neighborhoods $U,V$ of $x, C$ respectively.
Notice my definition of regularity doesn't ask $X$ to be $T_1$. I also understand that a regular space needs the $T_0$ axiom to be Hausdorff. My question is
Why is the $T_1$ axiom necessary in the proof of the following lemma?
Here's the lemma:
Lemma 31.1 Let $X$ be a $T_1$ topological space. Then $X$ is regular if and only if given a point $x\in X$ and a neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ such that $\overline V \subset U$ (where $\overline V$ denotes the closure of $V$)
Proof: Suppose $X$ is regular. Take any $x\in X$ and $U$ any of its neighborhoods. Then $U^c$ is closed and by hypothesis there are open disjoint neighborhoods $V, W$ of $x,U^c$ respectively. If $y\in \overline V\cap U^c$, then $y\in W$ but $W\cap V=\varnothing$ Hence $y\not\in\overline V$ and thus, $\overline V\cap U^c=\varnothing \Rightarrow \overline V\subset U$ as we wanted.
For the converse, take any $x\in X$ and a disjoint closed $C$. Then $C^c$ is an open neighborhood of $x$ and by hypothesis there is a neighborhood $V$ of $x$ such that $\overline V\subset C^c$. Then $V$ and $\overline V ^c$ are disjoint open neighborhoods of $x, C$ respectively. Hence $X$ is regular. $\blacksquare$
I understand all the steps but I don't see where the $T_1$ axiom hypothesis plays a key role. I'm pretty sure the reason to ask for the $T_1$ axiom is to avoid pathological cases, but I couldn't think of any.
Remember that a space is regular if $\textbf{all one-point set are closed}$ and each $\{x\}$ and closed subsets $A$ can be separated by disjoint open sets. The condition in the lemma is equivalent to the separation condition, but does not guarantee that the space is $T_1$. The fact that any pair $(\{x\},A)$ with $A$ closed, does not imply that the space is $T_1$.