Let $G$ be a Hausdorff topological group; let $X$ be a locally compact Hausdorff space; let $A$ be a C*-algebra.
Define an action of $G$ on $X$ to be a continuous map $G \times X \to X, \ (g,x) \mapsto gx$ such that $ghx = g(hx), \ ex = x$ for every $g,h \in G$; define an action of $G$ on $A$ to be a group homomorphism $G \to \text{Aut}(A), \ g \mapsto \alpha_g$ such that the map $g \mapsto \alpha_g(a)$ is continuous in the norm topology for every $a \in A$.
Given an action of $G$ on $X$, define the induced action of $G$ on $C_0(X)$ by $\alpha_g(f)(x) = f(g^{-1}x)$. It is clear that the induced action of $G$ on $C_0(X)$ is a group homomorphism from $G$ to the automorphisms of $C_0(X)$. However, how does one show that the map $g \mapsto \alpha_g(f)$ is norm continuous for every $f \in C_0(X)$?
One way to show the continuity of this map is as follows. Consider the basic open neighborhood $B_r(\alpha_g(f)) = \{F\in C_0(X) \mid \|F-\alpha_g(f)\|<r\}$ of $\alpha_g(f)$ in $C_0(X)$. Can we find a neighborhood $\mathcal N(g)$ of $g$ in $G$ such that when $h \in \mathcal N(g)$ we have $\alpha_h(f) \in B_r(\alpha_g(f))$?
Lemma 1: Let $K \subset X$ be compact and $U \subset X$ be open. If $g \in G$ satisfies $g(K) \subset U$, then there exists an open neighborhood $V \subset G$ of $g$ such that $h(K) \subset U$ whenever $h \in V$.
Proof: We note that the map $G \times X \ni (h, x) \mapsto hx \in X$ is jointly continuous. We denote this map by $\beta$. For any $x \in K$, by our assumption on $g$, $\beta^{-1}(U)$ contains $(g, x)$. Since $\beta^{-1}(U)$ is open, there exists $V_x \subset G$, open neighborhood of $g$, and $K_x \subset X$, open neighborhood of $x$, s.t. $\beta(V_x \times K_x) \subset U$. Observe that $K_x$ for $x \in K$ form an open cover of $K$. As $K$ is compact, there exists $x_1, x_2, \cdots, x_N \in K$ such that $K \subset \cup_{i=1}^N K_{x_i}$. Then $V = \cap_{i=1}^N V_{x_i}$ satisfies the desired condition. $\square$
Lemma 2: For any $K \subset X$ compact, there exists $U \subset X$ open such that $K \subset U$ and $\overline{U}$ is compact.
Proof: Since $X$ is locally compact, for each $x \in K$ there exists an open neighborhood $U_x$ of $x$ whose closure is compact. All such $U_x$ form an open cover of $K$, so there exists $x_1, x_2, \cdots, x_N \in K$ such that $K \subset \cup_{i=1}^N U_{x_i}$. Let $U = \cup_{i=1}^N U_{x_i}$. Then $U$ is an open neighborhood of $K$ whose closure is compact. $\square$
We now prove the result when $f$ is supported on a compact set $K \subset X$. Let $g \in G$, $\epsilon > 0$. Let $U$ be an open neighborhood of $g(K)$ whose closure is compact, which exists by Lemma 2. By Lemma 1, there exists $V_1 \subset G$, open neighborhood of $g$, s.t. $h(K) \subset U$ whenever $h \in V_1$.
Now, since $f$ is continuous, the map $G \times X \ni (h, x) \mapsto f(h^{-1}x) = \alpha_h(f)(x) \in \mathbb{C}$ is continuous. Thus, for each $x \in \overline{U}$, there exists open neighborhood $U_x \subset X$ of $x$ and an open neighborhood $V_x \subset G$ of $g$ s.t. $|\alpha_h(f)(y) - \alpha_g(f)(x)| \leq \epsilon/2$ whenever $y \in U_x$ and $h \in V_x$. All such $U_x$ form an open cover of $\overline{U}$, so as $\overline{U}$ is compact, there exists $x_1, \cdots, x_N \in \overline{U}$ s.t. $\overline{U} \subset \cup_{i=1}^N U_{x_i}$. Let $V_2 = \cap_{i=1}^N V_{x_i}$ and $V = V_1 \cap V_2$, which is an open neighborhood of $g$.
We claim that whenever $h \in V$, $||\alpha_h(f) - \alpha_g(f)|| \leq \epsilon$. For $x \notin U$, as $h, g \in V_1$, $h^{-1}x, g^{-1}x \notin K$, so $\alpha_h(f)(x) - \alpha_g(f)(x) = f(h^{-1}x) - f(g^{-1}x) = 0$ as $f$ is supported on $K$. If $x \in U$, then $x \in U_{x_i}$ for some $i$. Since $g, h \in V \subset V_{x_i}$, we have $|\alpha_h(f)(x) - \alpha_g(f)(x_i)| \leq \epsilon/2$ and $|\alpha_g(f)(x) - \alpha_g(f)(x_i)| \leq \epsilon/2$, so $|\alpha_h(f)(x) - \alpha_g(f)(x)| \leq \epsilon$. Hence, $|\alpha_h(f)(x) - \alpha_g(f)(x)| \leq \epsilon$ for all $x$, i.e., $||\alpha_h(f) - \alpha_g(f)|| \leq \epsilon$.
The above proves the result for compactly supported $f$. For the general result, let $f \in C_0(X)$ be arbitrary, $g \in G$, $\epsilon > 0$. Since compactly supported functions are dense in $C_0(X)$, there exists $f_0$, compactly supported, such that $||f - f_0|| \leq \epsilon/3$. By what we have shown, there exists an open neighborhood $V$ of $g$ such that $||\alpha_h(f_0) - \alpha_g(f_0)|| \leq \epsilon/3$ whenever $h \in V$. As $||f - f_0|| \leq \epsilon/3$, we clearly have $||\alpha_g(f) - \alpha_g(f_0)|| \leq \epsilon/3$ and $||\alpha_h(f) - \alpha_h(f_0)|| \leq \epsilon/3$, so whenever $h \in V$, we have $||\alpha_h(f) - \alpha_g(f)|| \leq \epsilon$. $\square$