Show that a non-identity analytic function has at most one fixed point

Question

Let $\hat{\mathbb{C}}$ be the extended complex plane, $G\subset\mathbb{C}$ be a proper open connected subset. Assume $\hat{\mathbb{C}}-G$ is connected. Show that an analytic function $f:G\to G$ which is not identity must have at most one fixed point.
I proved this as follows. First the assumption on connectedness of $\hat{\mathbb{C}}-G$ shows that $G$ is simply connected. Then use Riemann mapping theorem to reduce the problem to the case when $G$ is the unit disk, which can be solved using Schwartz lemma.
Since this is a qual exam problem and Riemann mapping theorem is not listed in the syllabus, I think it can be proved in a more elementary way without using Riemann mapping theorem. Is there such a proof?