My question is on page 163, the proof of Lemma 7 in the book Foundations of hyperbolic manifolds by John G. Ratcliffe.
Let $\Gamma$ be a group of isometries of a metric space $X$. If there is a point $x \in X$ such that the orbit $\Gamma x$ is discrete in $X$ and the stabilizer subgroup $\Gamma_x$ is finite, then $\Gamma$ is discrete.
He gives the following proof:
Let $\epsilon_x:\Gamma \to \Gamma x$ be the evaluation map. Then $\epsilon_x$ is continuous in the compact-open topology. Hence the set $\epsilon_x^{-1}(x) = \Gamma_x$ is open in $\Gamma$. Therefore, the identity map of $X$ is open in $\Gamma$, and so $\Gamma$ is discrete.
Now I don't understand why we get that the identity of $X$ is open in $\Gamma$.
Thank you for the help @LukasHeger I will try to finish the proof:
Since $\epsilon_{x}^{-1}(x)$ is open in $\Gamma$, its subspace topology is the same topology that it has as a subset of $\Gamma$. In particular, $\epsilon_{x}^{-1}(x)$ is Hausdorff and finite. And therefore discrete. That means that the identity $\{id\} \subset \Gamma_x$ must be open.