A space is regular if each closed set $Z$ is the intersection of all open sets containing $Z$? 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.
 A: Nate's example is a good one and actually solves the question as it is. But suppose you think that the Cromarty property, as Nate called it, is not good enough since 'almost all' spaces satisfy it. Then you try to substitute it for a stronger condition. You can require that every closed subset $Z \subseteq X$ is a countable intersection of open sets. That's exactly what a $G_{\delta}$ space is. In this situation the example Nate gave still works, you just need to require that the space $X$ is countable. In fact, if $F$ is a closed subset of $X$ with the cofinite topology then $X-F$ is the countable union of the closed sets $\{x\}$, with $x \in X - F$. So 
$$F = \bigcap_{x \in X-F}  (X - \{x\}).$$
Thus, a countable set $X$ with the cofinite topology is a $G_{\delta}$ space. But then you conclude that your condition isn't strong enough. Now you try to impose some kind of separation. The question becomes: is there a Hausdorff $G_{\delta}$ space that isn't regular?
The answer is yes. Let $K = \{ 1/n : n \in \mathbb{N} \}$ and consider the collection $\mathcal{B}$ of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$. It's easy to show that $\mathcal{B}$ is a basis for a topology on $\mathbb{R}$, which is called the K-topology. When $\mathbb{R}$ is endowed with this topology it's denoted by $\mathbb{R}_K$. 
The K-topology is finer than the standard topology on the real line. So it's a Hausdorff space (actually a completely Hausdorff space). Also, $\mathbb{R}_K$ is second countable, since the colletion $\mathcal{B'}$ of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$ and with $a,b \in \mathbb{Q}$ is a basis for the K-topology. $\mathbb{R}_K$ isn't regular, since the closed set and the point $0$ cannot be separated by disjoint open sets.
Now let's see why $\mathbb{R}_K$ is a $G_{\delta}$ space. It's easier to prove that it's an $F_{\sigma}$ space, that is, a space such that every open set can be written as a countable union of closed sets, which is equivalent to prove that it's a $G_{\delta}$ space. Since $\mathbb{R}_K$ is second countable, every open set can be written as a countable union of basis elements. If we prove that every basis element is an $F_{\sigma}$ set, then will follow that $\mathbb{R}_K$ is an $F_{\sigma}$ space.
The sets $(a,b)$ are $F_{\sigma}$, since $(a,b) = \bigcup_{n=1}^{\infty} [a + \frac{1}{n}, b - \frac{1}{n}]$. Now consider $(a,b) - K$; we do not need to worry with the end point $b$, so suppose $b > 1$. If $a>0$, let $n_0$ be the largest natural such that $a < 1/n_0$. Then 
$$ (a,b)-K = (a, \frac{1}{n_0}) \cup \bigcup_{n=1}^{n_0-1} (\frac{1}{n+1}, \frac{1}{n}) \cup (1,b).$$
Each of the sets in the decomposition above are $F_{\sigma}$, thus $ (a,b)-K$ is $F_{\sigma}$. If $a=0$, then 
$$ (0,b)-K = \bigcup_{n=1}^{\infty} (\frac{1}{n+1}, \frac{1}{n}) \cup (1,b),$$
so $(0,b)-K$ is $F_{\sigma}$. If $a<0$, then
$$ (a,b)-K = (a,0] \cup ((0,b)-K) = \bigcup_{n=1}^{\infty} [a + \frac{1}{n}, 0] \cup ((0,b)-K),$$
so $(a,b)-K$ is an $F_{\sigma}$ and we conclude that $\mathbb{R}_K$ is an $F_{\sigma}$ space.
A: 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 $R_0$.
A: Isn't the trivial (indiscrete) topology a counterexample?  Perhaps there are more assumptions somewhere.
Edit: I had a different definition of "regular" in mind from the asker.  The question has been clarified, and the trivial topology is not a counterexample.
