I'm in trouble show that for any sequence $\{x_n\}_n$ in a compact Riemann surface, there exist a subsequence $\{ x_{n_k}\}_k$ of $\{x_n\}_n$, a chart $(U,\varphi)$ and an $x\in U$ such that the sequence $\{\varphi (x_{n_k})\}_k$ converges to $\varphi(x)$.
Any help?
Since the Riemann surface is compact, you can cover it by finitely many charts $(U_i,\varphi_i)$, each of whose closures is compact. By the pigeonhole principle, find a chart $(U_{i_0},\varphi_{i_0})$ such that for infinitely many $n_k$, $x_{n_k}\in U_{i_0}$. Can you finish the proof from here?
The proof should make it clear that this is really not a proper statement about Riemann surfaces per se, since we have no use for the fact that the transition maps $\varphi_{UV}:\varphi_U(U)\longrightarrow\varphi_V(V)$ are holomorphic. It is at best a statement about compact and locally Euclidean topological spaces as this answer suggests.