Group action on isometric embeddings of compact set with uniform topology is continuous. This is taken from Vladimir Pestov's book.  
Let $\left(X, d_X \right)$ be a metric space and $K \subset X$ a compact subspace.
Now suppose a topological group $G$ acts continuously and isometrically on $X$.
Consider the space $X^{\hookleftarrow K}$ of all isometric embeddings of $K$ into $X$ with the topology given by the uniform metric  
\begin{equation}
d(i,j) = \sup \limits_{k \in K} \left \{ d_X( i(k), j(k) ) \right \}
\end{equation}
The above metric is well-defined because $K$ is compact and hence totally bounded.
Now we let $G$ act on $X^{\hookleftarrow K}$ by
\begin{equation}
(g.i)(k) = g.(i(k)) \qquad \forall g \in G, i \in X^{\hookleftarrow K}, k \in K
\end{equation}
How do I prove that this action is continuous?
My idea was to represent any nonempty open set $V \in X^{\hookleftarrow K}$ as a finite intersection of simpler open sets of the form
\begin{equation}
\mathcal{N}^{k,\epsilon,i} = \left \{ j \in X^{\hookleftarrow K} : d_X( i(k), j(k) ) < \epsilon \right \}
\end{equation}
(this is possible when $K$ is finite and I belive by a compactness argument for any compact $K$) and then to look at the preimages of these under the group action. Letting $\beta : G \times X^{\hookleftarrow K} \to X^{\hookleftarrow K}$ denote the action of $G$ on $X^{\hookleftarrow K}$, we have
\begin{align}
\beta^{-1} \left(\mathcal{N}^{k,\epsilon,i} \right) &= \left \{ (g, j) \in G \times X^{\hookleftarrow K} : d_X( i(k), g.j(k) ) < \epsilon \right \} \\
&= \left \{ (g, j) \in G \times X^{\hookleftarrow K} : d_X( g^{-1}.i(k), j(k) ) < \epsilon \right \}
\end{align}
from which we immediately get separate continuity of $\beta$.
How do I prove the joint continuity?
 A: For finite $K$, this turned out to be simpler than expected.  
First, denote left multiplication by $g \in G$ by $\lambda_g$. Then $\lambda_g$ is an open map for any $g \in G$.
Let $i \in X^{\hookleftarrow K}$, $g \in G$, $\epsilon > 0$ be arbitrary. 
By the continuity of the action of $G$ on $X$, the sets
\begin{equation}
H_{j,k}(\delta) = \left \{ g \in G : d_X \left( g.(j(k)), j(k) \right) < \delta \right \}
\end{equation}
where $j \in X^{\hookleftarrow K}$, $k \in K$ and $\delta > 0$, are open.
Because $K$ is finite, the set
\begin{equation}
H_j(\delta) = \left \{ g \in G : d \left( g.j, j \right) < \delta \right \} = \bigcap \limits_{k \in K} H_{j,k}(\delta)
\end{equation}
is also open.
Now pick any
\begin{equation}
(h, j) \in \lambda_g \left( H_{i} \left(\frac{\epsilon}{2} \right) \right) \times B_i \left(\frac{\epsilon}{2} \right)
\end{equation}
Then
\begin{equation}
d(g.i,h.j) \le d(g.i,h.i) + d(h.i,h.j) \le d \left(i, \left(g^{-1}h \right).i \right) + d(i,j) \le \frac{\epsilon}{2} + \frac{\epsilon}{2} \le \epsilon
\end{equation}
Hence there exists an open neighbourhood of $(g,i)$ that is mapped into $B_{g.i}(\epsilon)$ and this is true for any $i \in X^{\hookleftarrow K}$, $g \in G$ and $\epsilon > 0$.
This is equivalent to continuity by a standard argument (included below).
TLDR
For any $i \in X^{\hookleftarrow K}$ and $\epsilon > 0$ let
\begin{equation}
A_i(\epsilon) = \bigcup \limits_{(g,j) \in \beta^{-1}(i)} \lambda_g \left( H_j \left(\frac{\epsilon}{2} \right) \right) \times B_j \left(\frac{\epsilon}{2} \right)
\end{equation}
$A_i(\epsilon)$ is obviously open and by the above we have
\begin{equation}
\beta(A_i(\epsilon)) \subseteq B_i(\epsilon)
\end{equation}
Now let $U \subseteq X^{\hookleftarrow K}$ be open and pick any $i \in U$. Then there is an $\epsilon_i > 0 $ such that $B_i(\epsilon_i) \subseteq U$.
We then have
\begin{equation}
U = \bigcup \limits_{i \in U} B_i(\epsilon_i)
\end{equation}
But then
\begin{equation}
\beta \left( \bigcup \limits_{i \in U} A_i(\epsilon_i) \right) \subseteq \bigcup \limits_{i \in U} B_i(\epsilon_i) = U
\end{equation}
Now because for any $i \in U : \beta^{-1}(i) \subseteq A_i(\epsilon_i)$ we get that
\begin{equation}
\beta^{-1}(U) = \bigcup \limits_{i \in U} A_i(\epsilon_i)
\end{equation}
which is obviously open.
