Prove that the metric space of the isometries of a compact metric space, with uniform distance, is compact.
Formalizing (from my understanding):
Let $(M,d)$ be a compact metric space. Then take the set $I = \{f:M \rightarrow M \mid f \text{ is a isometry and } f \text{ is bounded}\}$. With that, $(I,d')$ where $d'(f,g) = \sup\{d(f(x),g(x)) \mid x \in M\}$ is the metric space of the isometries of $M$. We should then prove that $(I,d')$ is compact.
I know that in order to prove that $(I,d')$ is compact we must start with an open coverage of it and show that there is a finite sub coverage that covers our space. So my intuition tells me that the direction to go is to try to somehow from the open coverage of $(I,d')$ get a coverage for $M$, and using compactness of $M$ we'd finite sub coverage of that, which somehow would give us a finite open coverage for $(I,d')$?! Honestly, I'm lost in this exercise.
I'd like to see the proof of this, since I've spent a good amount of time trying to reach a solution and couldn't so far.
Any help is appreciated. Thanks!