Q:

The boundary of a unbounded simply connected planar domain

Let $D$ be a unbounded simply connected planar open domain. Let $\partial D$ be its boundary. The question is the following:
Can $\partial D$ have a compact component?

A:

Suppose that $D\subset R^n$ is an open unbounded connected subset (a domain) whose boundary has a compact connected component. Then $E=R^n\setminus D$ has at least 2 connected components. Now, it follows from the Alexander duality that $H_{n-1}(D)\cong \tilde{H}^0(E)$ (the right hand side is the Chech cohomology). Hence, $H_{n-1}(D)$ has rank at least 1. In your case $n=2$ and the Hurewicz theorem implies that $\pi_1(D)$ is nontrivial. Contradiction. 
In dimension 2 one can give a proof avoiding (co)homology, but I prefer this argument as more general. 

