Continuous kernel iff closed countable intersection of open sets [duplicate]

Suppose $S=\bigcap U_n$, where $U_n\subset X$ open. For each $n\in\mathbb N$, since $S\subset U_n$ is closed, we can apply Urysohn's lemma (this is where the assumption of normality is needed) to find continuous $f_n:X\to[0,1]$ s.t. $f_n=0$ on $S$ and $f_n=1$ on $X\setminus U_n$.

Now consider $f=\sum2^{-n}f_n$. By the Weierstrass M-test, we know that $f$ is continuous. Clearly $f=0$ on $S$, and for $x\not\in S$, we can find $n$ s.t. $x\not\in U_n$, so $f_n(x)\neq0$ and thus $f(x)\neq0$. Hence $S=\ker f$.

Let $S$ be closed, and $\langle O_n\rangle$ be a sequence of open sets whose intersection is $S$. Since we are in a normal space, we can use Urysohn's lemma to find for each $n$ a continuous function $f_n:X\to [0,1]$ that takes the value $0$ on the closed set $S$ and takes the value $1$ on the closed set $X\setminus O_n$. Then $f:X\to [0,1]$ given by $f(x)=\sum_{n=1}^\infty 1/2^n f_n(x)$ is a continuous function with kernel $S$.