$\newcommand{\cl}{\operatorname{cl}}$I have only the first edition, but he does not appear to have changed the argument much in the second edition. First let me recapitulate the construction. For each $r\in\Bbb Q$ you have an open set $U_r$. These open sets have the following properties:
- $A\subseteq U_0$;
- $U_1=X\setminus B$;
- $\cl U_p\subseteq U_q$ whenever $p<q$;
- $U_r=\varnothing$ if $r<0$; and
- $U_r=X$ if $r>1$.
For $x\in X$ you have $\Bbb Q(x)=\{r\in\Bbb Q:x\in U_r\}$, and $f$ has been defined by $f(x)=\inf\Bbb Q(x)$. We want to show first that if $x\in\cl U_r$, then $f(x)\le r$.
Suppose that $x\in\cl U_r$. If $s\in\Bbb Q$ and $s>r$, then by construction $\cl U_r\subseteq U_s$, so $x\in U_s$, and $s\in\Bbb Q(x)$. In other words, $\Bbb Q(x)\supseteq\{s\in\Bbb Q:s>r\}$. Thus, any lower bound for $\Bbb Q(x)$ is also a lower bound for $\{s\in\Bbb Q:s>r\}$. In particular, $\inf\Bbb Q(x)$, the greatest lower bound for $\Bbb Q(x)$, is a lower bound for $\{s\in\Bbb Q:s>r\}$: $\inf\Bbb Q(x)\le s$ for all rational $s>r$. Clearly $r$ is the greatest bound for $\{s\in\Bbb Q:s>r\}$, so $\inf\Bbb Q(x)\le r$. And $f(x)=\inf\Bbb Q(x)$, so $f(x)\le r$. $\dashv$
Now we want to show that if $x\notin\cl U_r$, then $f(x)\ge r$.
For each $s<r$ we know that $U_s\subseteq U_r$, so if $x\notin\cl U_r$, then $x\notin U_s$ for any $s<r$. In other words, if $s<r$, then $s\notin\Bbb Q(x)$. Take the contrapositive: if $s\in\Bbb Q(x)$, then $s\ge r$. That is, every member of $\Bbb Q(x)$ is at least as big as $r$, and $r$ is therefore a lower bound for $\Bbb Q(x)$. Clearly, then, $r\le\inf\Bbb Q(x)=f(x)$. $\dashv$
