I'm trying to understand the following proof from Walker:
Proposition. The zero-sets of $\beta X$ are countable intersections of closures in $\beta X$ of zero-sets of $X$.
Proof. If $Z$ is a zero set of $\beta X$, then it's the zero of some function $\beta f$ with $f \in C^*(X)$, $Z(\beta f)$. Thus: $$Z(\beta f)=\bigcap_{n=1}^\infty\{p \in \beta X: |\beta f(p)|\leq 1/n\}$$
But then we can write that:
$$Z(\beta f)=\bigcap_{n=1}^\infty cl_{\beta X}\{x \in X: |f(x)|\leq 1/n\}$$
Which completes the proof.
Question: I'm not seeing why does the first equation imples the second. I think that I must show that for each $n$ the equality $\{p \in \beta X: |\beta f(p)|\leq 1/n\}=cl_{\beta X}\{x \in X: |f(x)|\leq 1/n\}$ holds. $\supset$ is easy, but I'm having trouble with the other inclusion.
Indeed, the other inclusion does not necessarily hold. Consider e.g. for fixed $n$ the function $f_n\colon \mathbb{R} \to \mathbb{R}$ given by $f_n(x) = \frac{1}{n} + \frac{1}{1+x^2}$. Then $\{ x\in \mathbb{R} : \lvert f_n(x)\rvert \leqslant 1/n\} = \varnothing$, but $\{ p \in \beta\mathbb{R} : \lvert\beta f_n(p)\rvert \leqslant 1/n\} \neq \varnothing$.
However, since we take the intersection over all $n \in \mathbb{N}\setminus \{0\}$, it comes out correctly, since
$$\{ p \in \beta X : \lvert \beta f(p)\rvert \leqslant 1/n\} \subset \operatorname{cl}_{\beta X} \{ x\in X : \lvert f(x)\rvert \leqslant 1/(n-1)\}$$
for $n > 1$.
Thus we take an intersection of two interleaved nested (decreasing) sequences of sets, and then the intersections are equal.