 Solution

The bounded sequence $(x_n)$ lies in a compact set because $X$ is a proper metric space. After taking a subsequence, $x_n\to x$. Since $d(x_n,{\varphi }_n{x}_n)\to 0$, both $x_n$ and ${\varphi }_n{x}_n$ eventually lie in one compact neighborhood $L$ of $x$. A properly discontinuous group action has only finitely many $g\in \Gamma$ with $gL\cap L\ne \varnothing$, so some conjugate $\psi$ occurs as ${\varphi }_n$ along an infinite subsequence. Continuity then givesThus $\psi$ fixes $x$. Since $\psi =\gamma \varphi {\gamma }^{-1}$ for some $\gamma$, the original element $\varphi$ fixes ${\gamma }^{-1}x$.

Solved by gpt-5.6-sol high.