I want to use the uniformization theorem, which states that every simply connected riemann surface is biholomorphically equivalent to either the riemann sphere $\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$, the complex plane $\mathbb{C}$ or the open disc $D = \{ x \in \mathbb{C} : |x| < 1 \}$, to deduce that following corollary:
Corollary: Every simply connected domain $G \subseteq \hat{\mathbb{C}}$ which is missing at least two points from $\hat{\mathbb{C}}$ is biholomorphically equivalent to $D$.
It's clear to me that due to compactness $G$ cannot be biholomorphically equivalent to $\hat{\mathbb{C}}$. But how do I show it's not biholomorphically equivalent to $\mathbb{C}$? When $x_0 \neq x_1$ are points of $\hat{\mathbb{C}}$ not in $G$, we know that $\hat{\mathbb{C}} \setminus \{x_0\}$ is biholomorphic to $\mathbb{C}$. So $G$ can be identified with a subset of $\mathbb{C}$. Now how do I show that this subset cannot be biholomorphic to $\mathbb{C}$ without prooving the standard riemann mapping theorem on its own? (or do I have to maybe?)
Let $x\in\mathbb{C}$ and $G\subset\mathbb{C}\setminus\{x\}$ be simply connected. If we have a non-constant holomorphic map $f:\mathbb{C}\rightarrow G$, then $G=\mathbb{C}\setminus\{x\}$ by Picard's little theorem. Thus $f$ cannot be biholomorphic.