Continuity of a real function defined on an orbit space I am reading some paper, where they claim the following:
Let $G$ be a compact (Hausdorff) group and let $X$ be a locally, compact, Hausdorff space. Assume that $G$ acts on $X$ continuously. Denote by $C_0(X)$ the continuous functions on $X$ with complex values. Denote by $X/G$ the orbit space (it is Hausdorff as well), and let $\pi: X\to X/G$ be the canonical quotient map.
claim: Let $f\in C_0(X)$. The map $\pi(x)\mapsto sup_{g\in G}|f(g\cdot x)|$ is a continuous map $X/G\to \mathbb{R}$.
The proof starts like that: Let $f\in C_0(X)$, let $\epsilon>0$ and let $x\in X$. Use continuity of $f$ to find, for every $g\in G$, an open neighborhood $W_g$ of $g\cdot x$ such that $|f(g\cdot x)-f(y)|<\epsilon$ for all $y\in W_g$. By compactness of $G$, there exists an open neighborhood $W$ of $x$ such that $g\cdot W\subseteq W_g$ for all $g\in G$.
I can not see why the last claim is true or why it follows from compactness... it seems like they actually claim that $\bigcap\limits_{g\in G}g^{-1}\cdot W_g$ is open (of course, it contains $x$, but I don't see why should it be open...).
An alternative approach to the proof or counter example for the claim would be appreciated as well.
Thank for any help!
 A: As Clément Guérin suggested in the comments, a possible approach to prove this claim is by using Arzelà-Ascoli's Theorem in the generalised version for locally compact Hausdorff spaces, as it appears in Munker's book for topology. Can be found also here: Theorem of Arzelà-Ascoli.
With $f$, $x$ and $\epsilon>0$ as above, we have the family $\mathcal{F}=\{g\cdot f| g\in G\}$ of continuous functions (where I realise the action on $X$ as an action of $C_0(X)$, naturally). Since $G$ is compact and $\mathcal{F}$ is just the orbit of $f$ under the action, we get that $\mathcal{F}\subseteq C_0(X)$ is a compact subset. Therefore, we can apply Arzelà-Ascoli Theorem and obtain that this family is equicontinuous: There exists an open neighborhood $W\subseteq X$ of $x$ such that $|f(g\cdot x)-f(g\cdot y)|<\epsilon$ for all $g\in G$ and all $y\in W$.
Since the quotient map $\pi$ is open, $U:=\pi(W)$ is an open neighborhood of $\pi(x)$ in $X/G$. By the above, it follows that for all $y\in W$:
$sup_{h\in G}|f(h\cdot y)|-\epsilon\leq sup_{g\in G}|f(g\cdot x)|\leq sup_{h\in G}|f(h\cdot y)|+\epsilon$ , and therefore:
$sup_{\pi(y)\in U}|sup_{h\in G}|f(h\cdot y)|-sup_{g\in G}|f(g\cdot x)||<\epsilon$, as required.
