Is a closed $G_\delta$ set in a Hausdorff space always a zero set? I have been trying to prove that if $A$ is a closed set which is also an intersection of countably many open sets then $A$ is the zero set for some continuous real-valued function however have thus far failed. Is this even true?
 A: It is not true.
John Thomas, A regular space, not completely regular, Amer. Math. Monthly 76 (1969), 181-182, constructed a regular Hausdorff space $X$ with two points, $p$, and $q$, such that for each continuous $f:X\to\Bbb R$, $f(a)=f(b)$. Moreover, $X$ has countable local bases at $p$ and $q$. Thus, $\{p\}$ is a closed $G_\delta$-set in $X$ that cannot be a zero-set: any zero-set containing $p$ must also contain $q$.
A. Mysior, A regular space which is not completely regular, Proc. Amer. Math. Soc. 81 (1981), 652-653, is freely available and has a regular Hausdorff space with a point $p$ and a closed set $F$ such that for each continuous $f:X\to\Bbb R$ and $x\in F$, $f(x)=f(p)$. The point $p$ has a countable local base, so here again $\{p\}$ is a closed $G_\delta$ that cannot be a zero-set.
A: Not even in completely regular Hausdorff spaces.  In general we have
$$
\text{compact $G_\delta$}\qquad\Longrightarrow\qquad
\text{zero-set}\qquad\Longrightarrow\qquad
\text{closed $G_\delta$}
$$
but none reversible.
A: Brian's answer covers the question fully. For fun, here's another example:
Bing's irrational slope space is a countable and connected Hausdorff space. 
Now observe:


*

*If $f \colon X \to \mathbb{R}$ is continuous, then $f(X)$ is a countable and connected subset of $\mathbb{R}$, hence it must be reduced to a point. Therefore all continuous functions $f \colon X \to \mathbb{R}$ are constant, and the only zero sets are the empty set and the space itself.

*Since $X$ is countable and $T_1$, every subset $F$ of $X$ is a $G_\delta$-set: $F = \bigcap_{x \in X \setminus F} X \setminus \{x\}$, in particular, there is an abundance of closed $G_\delta$ sets that are not zero sets.
