Showing that a simply connected proper subset of $\mathbb{C}$ is not conformally equivalent to $\mathbb{C}$

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So far, I have said that if $U$ is proper, open and simply connected, then its complement contains an open set. Suppose $f$ is a conformal equivalence form $\mathbb{C}$ tp $U$. Consider the induced meromorphic function on the Riemann sphere. By Casorati-Weierstrass, since the image does not include the set missed in $\mathbb{C}-U$, so the singularity at infinity is a pole. Since $f$ is bijective and analytic, then by the Valency Theorem the pole must be simple.

I don't know where to go from here.