I'm trying to prove that all hyperbolic cyclic subgroups of $PSL(2, \mathbb{R})$ are Fuchsian.
Let $\Gamma$ be a hyperbolic subgroup of $PSL(2, \mathbb{R})$. It is enough to show that $\Gamma$ is discrete, since all discrete subgroups of $PSL(2, \mathbb{R})$ are Fuchsian.
My aim was to do this by contradiction.
Assume $\Gamma$ is not Fuchsian. This implies there exists a sequence $\{T_k\}$ with $T_k \in \Gamma$ and $T_k \rightarrow Id$ as $k \rightarrow \infty$.
From here I don't know where to go, but surely I will have to use the fact that every hyperbolic element $T \in\Gamma$ has two fixedpoints, say $z_1, z_2$.
You have the right idea by finding some sequence $\{T_k\}$ which converges to the identity. How to do this, you have to recall that we can conjugate any hyperbolic elements to $z\mapsto \lambda z$ for $\lambda≠1$.
For more detail see my answer here.