Is every $G_\delta $, zero-set? Zero-set means a set of the form:
$Z(f) = \{ x \in X | f(x) = 0 \}\quad\text{for some } f \in C(X)$
$C(X)$ is the ring of continuous function on $X$.
I know that  every zero-set is  $G_\delta $, i.e, a countably intersection of open  sets.

Is every $G_\delta $, zero-set? if not, can you give me an simple example.

 A: Trivially not, in general: there is no continuous function on $\mathbb{R}$ whose zero set is $(0,1)$, since if $f$ were such a function, then it would satisfy $$f(1) = \lim_{x\to1}f(x) = \lim_{x\to1^-}0 = 0,$$ so $1 \in Z(f)$, so $Z(f) \neq (0,1)$, but as an open set, $(0,1)$ is clearly $G_\delta$: it is the intersection of countably many copies of itself. 
A: $Z(f)$ has to be closed, so any $G_\delta$ set which is not closed will be a counterexample.
As a very specific example, in $\mathbb{R}$, the set $(0,1)$ is $G_\delta$ (indeed it's open) but it cannot be the zero set of any continuous function.
It's also not true in general that every closed $G_\delta$ set is a zero set, not even if $X$ is Hausdorff.  See Is a closed $G_\delta$ set in a Hausdorff space always a zero set?.  (It is true if $X$ is $T_6$, and maybe some weaker condition would also suffice.)
A: A zero-set of $X$ is always a closed $G_\delta$ by construction, in any $X$.
A closed $G_\delta$ set is a zero-set when $X$ is metric or more generally $T_4$.
This need not hold in all spaces, though. You'll meet counterexamples naturally later in the study of rings of continuous functions.
