Theorem: Let $(X,d)$ be a complete metric space and suppose that $\emptyset \ne \mathcal{F}\subset \mathcal{C}(X,\mathbb{K})$. Suppose that for each $x\in X$ there exists $\kappa_x>0$ such that $|f(x)|\le \kappa_x$ for all $f\in\mathcal{F}$. Then there exists a non-empty open set $G\subset X$ and $\kappa>0$ such that $|f(x)|\le \kappa$ for all $x\in G$ and for all $f\in\mathcal{F}$.
Proof: For each $m\ge 1$, let $$H_{m,f}:=\{x\in X: |f(x)|\le m\},$$ and let $$H_m := \bigcap_{f\in\mathcal{F}} H_{m,f}.$$ Since each $f\in\mathcal{F}$ is continuous, each $H_{m,f}$ is closed, and thus $H_m$ is closed. Moreover, for each $x\in X$, there exists $m\ge 1$ such that $|f(x)|\le m$ for all $f\in\mathcal{F}$, and so there exists $m\ge 1$ such that $x\in H_m$. That is, $X=\bigcup_m H_m$. But $X$ is complete, so by the Baire Category Theorem, at least one of the sets $H_m$ is not nowhere dense. That is, there exists $N\ge 1$ such that $G:=\mbox{int }H_N\ne \emptyset$. Let $\kappa = N$ to get that for $x\in G$, $$|f(x)|\le \kappa \mbox{ for all $f\in \mathcal{F}$}.$$
I would appreciate if someone could please explain to me how it is deduced that if $G$ is not nowhere dense then $|f(x)|\le N$ for all $x\in G$. It seems to me that one of the intermediary steps [explanations] is missing from this proof.