Let $(X,d)$ be a metric space and let $\mathrm{Iso}(X)$ denote the group of isometries of $(X,d)$, where the group operation is the composition. Equip $\mathrm{Iso}(X)$ with the topology of pointwise convergence. I have not been able to find any source supporting the claim that $\mathrm{Iso}(X)$ is a topological group. Yet, it seems to me that this claim is really true:
Since the topology on $\mathrm{Iso}(X)$ is the topology of pointwise convergence, it suffices to show that, given any $x \in X$, the mappings $f \mapsto f^{-1}(x)$ and $(f,g) \mapsto f(g(x))$ are continuous. To show the continuity of the mapping $f \mapsto f^{-1}(x)$, let $f \in \mathrm{Iso}(X)$ and $\varepsilon >0$ be given. Define $\mathcal{U}:= \big\lbrace g \in \mathrm{Iso}(X) \, ; \ g(f^{-1}(x))\in B(x,\varepsilon) \big\rbrace$, where $B(x,\varepsilon)=\{ y \in X \, ; \ d(x,y)<\varepsilon \}$. Then $\mathcal{U}$ is an open set, it contains $f$ and, for every $g \in \mathcal{U}$, we have $d\big(f^{-1}(x),g^{-1}(x)\big)=d \big( g(f^{-1}(x),x) \big)<\varepsilon$.
To show that $(f,g) \mapsto f(g(x))$ is continuous, let $f,g\in \mathrm{Iso}(X)$ and $\varepsilon >0$ be given. Let $$\mathcal{U}:= \big\lbrace u \in \mathrm{Iso}(X) \, ; \ u(g(x))\in B\big(f(g(x)),\varepsilon/2\big) \big\rbrace ,$$ $$\mathcal{V}:= \big\lbrace v \in \mathrm{Iso}(X) \, ; \ v(x)\in B(g(x),\varepsilon/2) \big\rbrace .$$ Then $\mathcal{U}$ and $\mathcal{V}$ are open sets, they contain $f$ and $g$, respectively, and, for all $u \in \mathcal{U}$ and $v \in \mathcal{V}$, we have $$d \big( f(g(x)),u(v(x)) \big) \leq d \big( f(g(x)),u(g(x)) \big) + d\big( u(g(x)),u(v(x)) \big) = d \big( f(g(x)),u(g(x)) \big) + d \big( g(x),v(x) \big) < \frac{\varepsilon}{2}+\frac{\varepsilon}{2}.$$
Is my proof correct? Am I right that $\mathrm{Iso}(X)$ is a topological group? If yes, I don't understand why this is not considered one of basic examples of a topological group. I would expect the Wikipedia article on topological groups to mention it.