I'm looking for easy proofs (or just an easy proof) of the following statement:
Let X be a hyperbolic Riemann surface, i.e., $X$ is a Riemann surface and the universal covering of $X$ is the complex upper half plane. Then the fundamental group of X is non-abelian.
One could resort to several different proofs:
Compute the fundamental group. This is a standard computation. (This is too "difficult" though.)
Use that X is algebraic and the complex upper half plane isn't. Therefore, its fundamental group is an infinite subgroup of PSL$_2(\mathbf R)$.