I've been trying to prove the following statement, but the converse has been giving me trouble.
A topological space $X$ is regular if and only if every closed subset $Z\subseteq X$ is the intersection of all open sets $U\subseteq X$ which contain it.
Here, the defintion of regular is that for any point $p\in X$ and any closed subset $Z\subset X$ not containing $p$, there exist open sets $E,F\subset X$ such that $p\in E$, $Z\subseteq F$, and $E\cap F=\emptyset$. The space does not necessarily have to be a $T_1$ space.
I believe I showed the forward statement correctly. I suppose $X$ is regular, and clearly $Z\subseteq\cap\mathcal{U}$, where $\mathcal{U}$ is the family of all open sets containing $Z$. Then if I take $p\in\cap\mathcal{U}$, then $p$ is in every open subset containing $Z$. Towards a contradiction, I assume that $p$ is not in $Z$. Then since $X$ is regular, there exist open sets $E,F\subset X$ such that $p\in E$, $Z\subseteq F$, and $E\cap F=\emptyset$. Thus $p\not\in F$, but $F$ is an open set containing $Z$, and thus must contain $p$, a contradiction. Thus $p\in Z$, and so $Z=\cap\mathcal{U}$.
The backwards step has stumped me for a while. I take any point $p$ and a closed set $Z$ which does not contain $p$. So $Z=\cap\mathcal{U}$, and since $p\not\in Z$, there exists an open set $F$ containing $Z$ which does not contain $p$. The only observation I've been able to make is that $F^c$ is a closed set containing $p$, and then $F^c=\cap\mathcal{W}$ where $\mathcal{W}$ is every open set containing $F^c$. Hence $p$ is in every open set containing $F^c$. Taking any open sets $E$ and $F$ such that $p\in E$ and $Z\subseteq F$, and I haven't been able to see a way that there exists two such open sets that are disjoint, to show the space is regular. Showing $F^c$ is an open set would work, but I'm not sure if that's even true. I was hoping someone could point out what I'm missing, thank you.
Best Answer
Ok, let's try another counterexample.
For simplicity, say a space with your property (any closed set is the intersection of its neighborhoods) is "Cromarty".
Let $X$ be an infinite set with the cofinite topology (the closed sets are all the finite sets and $X$ itself). If $Z \subset X$ is closed (hence finite), let $\mathcal{U}$ be the collection of all open sets containing $Z$. For any $x \notin Z$, $\{x\}^c$ is open (because it is cofinite) and contains $Z$, so $\{x\}^c \in \mathcal{U}$, and thus $x \notin \bigcap \mathcal{U}$; hence $Z^C \subset \bigcap \mathcal{U}$, so $\bigcap \mathcal{U}=Z$. So $X$ is Cromarty.
However, $X$ is not regular, since every pair of nonempty open sets has nonempty (in fact, infinite) intersection.
Edit: To amplify, it looks like the Cromarty property is equivalent to being $R_0$, which means: if $x$ has a neighborhood not containing $y$, then $y$ has a neighborhood not containing $x$. This is of course strictly weaker than being regular (as the cofinite topology shows).
Suppose $X$ is $R_0$, and $Z$ is closed. Suppose $x \notin Z$, $y \in Z$. Then $Z^c$ is a neighborhood of $x$ not containing $y$, so $y$ has a neighborhood $U_y$ that does not contain $x$. Now $U = \bigcup_{y \in Z} U_y$ is a neighborhood of $Z$ that does not contain $x$. As above, it follows that $Z$ is the intersection of its neighborhoods, so $X$ is Cromarty.
Conversely, suppose $X$ is Cromarty. Suppose a point $x$ has a neighborhood $U$ not containing $y$. $U^c$ is closed, hence the intersection of its neighborhoods, so $U^c$ has a neighborhood $V$ not containing $x$. But $V$ is also a neighborhood of $y$, so $X$ is $R_0$.