Let $X$ be a compact Riemann surface and $f$ a nonconstant meromorphic function on $X$. Show that $f$ must have a zero on $X$, and must have a pole on $X$.
Suppose that $f$ is a nonconstant meromorphic function on a compact Riemann surface $X$. Then the associated mapping $F : X \to \mathbb{C}_{\infty}$ is a nonconstant holomorphic map. By Liouville's theorem, $F$ is subsequently not bounded. Hence, by the way in which $F$ is defined, $f$ has a pole since $F$ attains $\infty$. For $f$ meromorphic on a compact Riemann surface however, $$\sum_p \text{ord}_p(f) =0.$$ Therefore, $f$ must also have a zero.
This proof is wrong however, because I have clearly misused Liouville's theorem.
Let $f$ be a non-constant meromorphic function on a compact Riemann surface $X$. Consider the associated (non-constant) holomorphic mapping $F\colon X\rightarrow \mathbb{C}_\infty$.
By the Open Mapping Theorem (which can be derived from the Local Normal form of a non-constant holomorphic map between Riemann surfaces), the image $F(X)$ is open. Additionally, since $F$ is continuous and $X$ is compact, it follows that $F(X)$ is compact. Now, compact subsets of a Hausdorff space are closed. Thus, the non-empty set $F(X)$ is clopen. Since $\mathbb{C}_\infty$ is connected, $F$ is surjective. In particular, $f$ has both a zero and a pole.
By the way, this argument can be used to give a neat proof of the Fundamental Theorem of Algebra.