Let $(X, d)$ be a complete metric space and $A$ a set of continuous functions $f:X\to \mathbb{R}$ s.t. the sets $S_x = \{f(x)\mid \forall f \in A\}$ are bounded for each $x \in X$. Define $P_n = \{x\in X\mid |f(x)| \leq n, \forall f \in A\}$. Take it as granted that all $P_n$ is a closed set for all $n \in \mathbb{N}$.
I would like to show that $P_n$s also have empty interiors, i.e. no point in $P_n$ has a neighborhood contained in $P_n$. As sad as it may sound, I have no clue how to start the proof. Namely, suppose otherwise that $x \in \mathrm{int}(P_n), P_n \neq \varnothing$. Then for some $r > 0$ it holds true that the open ball $B(x, r)$ is contained in $P_n$. And then what? I don't really see how the closedness of $P_n$ can help us, so maybe the continuity of each $f \in A$ or the assumption regarding $A$ could be utilized? Well since $B(x, r) \subset P_n$, it follows that $f[B(x ,r)] \leq n, \forall f \in A$, so not really helpful, and I don't see immediately how the continuity of any $f \in A$ is helpful.
So how should I proceed? Any hints/tips regarding this problem are welcome!