Closed G-invariant set in a metric space contained in an open G-invariant set?

Question

Let $G$ be a countable discrete group and $X$ a metric space on which $G$ acts continuously and isometrically. Let $V$ be a closed $G$-invariant set and $U$ be an open $G$-invariant set such that $V \subset U$. There exists an open set $U^{\prime}$ such that $V \subset U^{\prime} \subset \overline{U^{\prime}} \subset U$ since $X$ is normal.

My question is whether $U^{\prime}$ can be taken to be $G$-invariant. I've tried various proof attempts and can see how to prove it if $G$ is finite (take the intersection of $gU^{\prime}$ over all $g \in G$ where $U^{\prime}$ is some open set containing $V$) but am unsure if this holds when $G$ is infinite. Thanks in advance.

Answer 1

Here is an actual proof. We have two disjoint closed $G$-invariant subsets of $X$: $V$ and $U^c=X\setminus U$. The functions $f, h: X\to {\mathbb R}$, $f(x)=d(x, V)$, $h(x)=d(x, U^c)$ are continuous and $G$-invariant. Define $$ U'= \left\{ x: f(x) - h(x)< 0 \right\}. $$
This is an open $G$-invariant subset containing $V$. Its closure is contained in $$ \left\{x: f(x) - h(x)\le 0\right\}, $$ which is disjoint from $U^c$. Hence, $cl(U')\subset U$.

Answer 2

Thanks to ChadK for the help.

Let $U^{\prime}$ be any open set containing $V$ such that $\overline{U^{\prime}} \subseteq U$. Then $V = gV \subseteq gU^{\prime}$ for all $g$. Since $G$ acts isometrically, there exists $\delta > 0$ such that $d(U^{\prime},U^{C}) > \delta$. Then $d(gU^{\prime},U^{C}) = d(gU^{\prime},gU^{C}) = d(U^{\prime},U^{C}) > \delta$ so $d(\cup gU^{\prime},U^{C}) > \delta$. Set $U_{0} = \cup gU^{\prime}$. Then $U_{0}$ is open and $G$-invariant. Since $d(U_{0},U^{C}) > \delta$, then $d(\overline{U_{0}},U^{C}) \geq \delta$. Therefore $V \subseteq U_{0} \subseteq \overline{U_{0}} \subseteq U$.