Let $A$ be a closed subset of a normal space $X$.
Show that there exists a continuous function $f:X\rightarrow[0,1]$ such that:
1) $\forall a\in A\; :\;f(a)=0$
2) $\forall x\in X-A\; :\;f(x)>0$
if and only if $A$ is a countable intersection of open sets in $X$.
The header might be misleading, but my best guess is that it is somehow related to Urysohn's lemma. I managed to prove the forward direction by looking at the sequence of closed intervals $$B_n=\left[ \frac{1}{n},1\right]$$ for which every $f^{-1}(B_n)$ is closed by continuity of $f$ and disjoint to $A$, thus from normality of $X$ separable by a pair of open and disjoint sets $U_n,V_n$. Then the intersection of all $U_n$'s yields $A$.
But for the reverse direction I'm not sure which way to go. Constructing an explicit function seems harder than using theory to ensure existence, and the only theory I can think of that relates normal spaces and continuous functions in such a manner is Urysohn's lemma. But I can't seem to get it to work...
Suppose $A = \cap_n U_n$ where all $U_n$ are open. Define for each $n$: $F_n= X\setminus U_n$. Then $A$ and $F_n$ are disjoint and closed sets in $X$, so we can find an Urysohn function $f_n : X \to [0,1]$ such that $$\forall x \in A: f_n(x) = 0 \land \forall x \in F_n: f_n(x) = 1$$
Now define $$f: X \to [0,1] \text{ by } f(x) =\sum_n \frac{1}{2^n}f_n(x)$$
Then $f$ is continuous (the series uniformly converges everywhere on $X$ )
And $f(x) = 0$ for all $x \in A$ and if $x \notin A$, then $x \notin U_m$ for some $m$, and then $f(x) \ge \frac{1}{2^m}f_m(x)$ (as all terms are positive) and this equals (as $x \in F_m$) $\frac{1}{2^m} > 0$.