Is a compactly supported function on a locally compact Hausdorff space uniformly continuous? Consider the following fragment from Folland's book "A course in abstract harmonic analysis":
Let $G$ be a locally compact Hausdorff group.

I want to prove that the representation
$$\pi: G \to B(L^2(\mu))$$
defined by $(\pi(x)f)(s) = f(x^{-1}s)$ is strongly continuous. I.e. if $f \in L^2(\mu)$ and $\{g_\alpha\} \to 1$, we must show that
$$\int_G |f(g_\alpha^{-1}s)-f(s)|^2 d \mu(s) = \|\pi(g_\alpha)f-f\|_2^2 \to 0.$$
My attempt: A routine density argument shows that it is sufficient to prove this for $f \in C_c(S)$.
In the hope of accomplishing this, I try to prove the following lemma (I try to mimique the proof of proposition 2.42, as suggested by Folland):

Lemma: Let $X$ be a locally compact Hausdorff space and $G$ a locally compact Hausdorff topological group together with a continuous
action $G\times X \to X$. If $f \in C_c(X)$, then the following
property is satisfied:
For every $\epsilon >0$, there exists an open subset $V\subseteq G$
containing $1$ such that for all $g \in V$, we have $\sup_{x \in
 X}|f(g^{-1}x)-f(x)| <\epsilon$.

Given this lemma, I can finish, but the classical proof (where $X=G$ and we consider the translation action) does not generalise to general actions.
To prove the lemma, I tried to modify the proof of proposition 2.6 in Folland's book, but it becomes clear that this proof no longer works, though I still hope that there exists some modification that saves the argument.
Any help/hint/suggestion/answer is highly appreciated! Thanks in advance for any help!
 A: Here is a proof of the Lemma in the OP based an a nice auxiliary result by Lynn H. Loomis.

Lemma (Loomis): Suppose $X,Y$ are locally compact Hausdorff (l.c.H) spaces and $Z$ a topological space, and let $f:X\times Y\rightarrow Z$ be a  continuous function. If $K\subset X$ is compact and $U\subset Z$ open, then
$$W=\{y\in Y: f(x,y)\in U\ \text{for all}\ x\in K\}$$
is open in $Y$.

Proof:  Fix $y_0\in W$. We show that there is an open neighborhood $V_{y_0}$ of $y_0$ fully contained in $W$. The continuity of $f$ implies that for each $x\in K$, there are open neighborhoods $A_x\subset X$ containing $x$ and $B_x$ containing $y_0$ such that $f(A_x\times B_x)\subset U$. By compactness of $K$, there are finite $x_1,\ldots, x_m\in K$ such that $K\subset \bigcup^m_{j=1}A_{x_j}$. The set $V_{y_0}=\bigcap^m_{j=1}B_{x_j}$ is open, contains $y_0$. Suppose $y\in V_{y_0}$,  and  let $x\in K$. Choose $A_{x_j}$ containing $x$. It follows that $(x,y)\in A_{x_j}\times V_{y_0}\subset A_{x_j}\times B_{x_j}$ and so, $f(x,y)\in U$.

As in the OP, let $X$ be a l.c.H space and $G$ a l.c.H topological group that acts continuous on $X$, and let $f\in\mathcal{C}_c(X)$, ale let $K=\operatorname{supp}(f)$. Recall that an action can be seen as a continuous function $\Phi:G\times X\rightarrow X$ such that $\Phi(1,x)=x$ for all $x\in X$ and  $\Phi(gh,x)=\Phi(g,\Phi(h, x))$ for all $g,h\in G$ and $x\in X$. It is usual to denote the group action as $g\cdot x:= \Phi(g,x)$.
Choose any symmetric neighborhood $U$ of $1\in G$ with compact closure. Define $F:X\times G\rightarrow \mathbb{R}$ by
$$F(x,g)= f(g^{-1}\cdot x)-f(x)=f(\Phi(g^{-1},x))-f(x)$$
This is a continuous function. Notice that  $\overline{U}\cdot K:=\Phi(\overline{U}\times K)\subset X$ is compact. The Lemma above shows that
$$W=\{g\in G: |F(x,g)|<\varepsilon,\,\text{for all}\, x\in \overline{U}\cdot K\}$$
is an open subset of $G$ which contains $1$; hence, $V:=(W\cap W^{-1})\cap U$ is a symmetric open neighborhood of $1$.
Let $s\in V$, and let $x\in X$. If $x\in \overline{U}\cdot X$, then by definition of $W$,
$$|f(s^{-1}\cdot x)-f(x)|=|F(x,s)|<\varepsilon$$
If $x\notin \overline{U}\cdot K$, then $g^{-1}\cdot x\notin K$  for all $g\in V$. Indeed, if $g^{-1}\cdot x\in K$ for some $g\in V$, then  $x=ss^{-1}\cdot x=s\cdot(s^{-1}\cdot x)\in U\cdot K$ which is a contradiction.  This means that
$$|f(s^{-1}\cdot x)-f(x)|=|f(s^{-1}\cdot x)-f(1\cdot x)|=0<\varepsilon$$
This proves the Lemma in the OP.
