In summary: Since A is closed and X is normal, and since X is an open neighborhood of A, we can find an open neighborhood U1 of A such that Cl(U1) is contained in X. Further on, for U1 and A, we can find an open neighborhood U2 of A such that Cl(U2) is contained in U1, etc. Now, from this we can conclude that A is contained in the countable intersection of open neighborhoods U1, U2, etc. This is correct. However, I feel obliged to tell you that there is an easier proof. It is so that A=f^{-1}(\{0\}). And the inverse of a closed set
Here, for convenience, I'll just sum up the sketch of the proof of the other direction.
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Let A be a closed Gδ set in X, so it equals a countable intersection of open sets in X. For any positive integer m, define Am = [tex]\cap_{i=1}^m A_{i}[/tex]. Further on, for any positive integer m, apply the Urysohn lemma to the disjoint closed sets A and X\Am in order to obtain a continuous function fm such that fm(A) = 0 and fm(X\Am) = 1. So, we obtain a sequence of continuous functions {fm}.
Now, for every positive integer m, define Sm = {x in Am\A : fm(x) = 0}. Then fm^-1({0}) = A U Sm[tex]\subseteq[/tex]Am, for every m. Now, [tex]\cap_{m}[/tex](A U Sm) = A U ([tex]\cap_{m}[/tex]Sm) = A, which implies [tex]\cap_{m}[/tex]Sm = [tex]\emptyset[/tex]. So, for any x in X, there exists some positive integer m such that fm(x)[tex]\neq[/tex]0. This motivates the following definition:
Define h(x) = [tex]\sum_{k=1}^{\infty}\frac{1}{2^k}f_{k}(x)[/tex]. It is easily checked that for any x in A, h(x) = 0, and for any x in X\A, h(x) > 0. Also, one can show that this series converges uniformly, and that its sum (i.e. h(x)) is a continuous function.
