A topological space is regular if and only if any closed set $Z$ is the intersection of its closed neighborhoods?

This is a follow up to a question I asked earlier. I had previously asked that if a space was regular if and only if any closed set $Z$ was the intersection of all open sets containing $Z$, but the backwards implication turned out to be untrue. I'm now curious to see if this new bi-implication is fully true.

My definition of regular here is that for any $p\in X$ and any closed set $Q$ not containing $p$, there exist open sets $U,V$ such that $p\in U$, $Q\subseteq V$, and $U\cap V=\emptyset$.

I assume $X$ is regular. I think it is clear that $$Z\subseteq\bigcap_{\text{closed}\ N\in\mathscr{N}_Z}N$$

For the other containment, I suppose that $p\not\in Z$. Then since $X$ is regular, there exist open sets $U,V$ such that $p\in U$, $Z\subseteq V$, and $U\cap V=\emptyset$. Then $U\subseteq V^c\subseteq Z^c$, and thus $U^c\supseteq V\supseteq Z$. So $U^c$ is a closed neighborhood of $Z$ which does not contain $p$. Hence $p\not\in\cap\mathcal{N}$, where $\mathcal{N}$ is the family of closed neighborhood of $Z$, and the equality follows.

I feel I have a hole in my reasoning to show the other implication. I suppose now that $Z=\cap\mathcal{N}$, and take some $p\not\in Z$, and so $p\not\in\cap\mathcal{N}$. Anyway, $p\in Z^c$, which is an open neighborhood of $p$. I then take some closed neighborhood $Q$ of $p$ such that $Q\subseteq Z^c$. I then let $U={Q}^{\circ}$, and $V=Q^c$. So both are are open and disjoint. Also, $p\in U$, and $V=Q^c\supseteq (Z^c)^c=Z$, so $X$ is regular.

However, I don't think I'm able to make the assumption that there is some closed neighborhood of $p$ contained in the open neighborhood $Z^c$. Is there some way to use the fact that $p\not\in\cap\mathcal{N}$ to show this is indeed true? Thanks.

EDIT: In case it is not entirely clear, my definition of a neighborhood $N$ of a set $Z$ is that $Z$ is contained in the interior of $N$, so $N$ is a neighborhood of each point $Z$. Then my definition of a neighborhood $M$ of a point $p$ is that $M$ contains an open set containing $p$. $\mathscr{N}_Z$ is my notation for the set of all neighborhoods of $Z$, and $\mathcal{N}$ is a shortcut way of writing the set of $\text{closed}\ N\in\mathscr{N}_Z$.

I agree that you don't know that there is a closed neighborhood of $p$ contained in $Z^c$.
However, this perhaps will work: Since $p$ is not in the intersection $\cap \mathcal{N}$, there is a closed set $V$ with $Z\subseteq V^{\circ}$ and with $p\notin V$. Since $V$ is closed, then $V^c$ is an open set containing $p$; and the interior of $V$ is an open set containing $Z$. Of course, $V^c$ is disjoint from $V^{\circ}$, so don't $V^{\circ}$ and $V^c$ give you open sets, disjoint, one containing $Z$ and the other containing the point $p\notin Z$?
• @Cromarty: You were indeed very close; your error was staring to closely at $Z$, instead of looking at one those closed neighborhoods thingies. (-; – Arturo Magidin Nov 24 '10 at 2:20