We have a simply connected domain $D$ in the Riemann sphere $\hat{\mathbb{C}}$ such that the complement $D \backslash \hat{\mathbb{C}}$ has more than one point. We use a Mobius transform $g$ to send one point of the complement to $0$ and one to $\infty$. Now we know that $g(D)$ is connected so that there is an arc $\alpha$ from $0$ to $\infty$ that lies outside of it.
The next step confuses me.
We can choose an inverse branch $h:\hat{\mathbb{C}}\backslash \alpha \rightarrow\hat{\mathbb{C}} $ of the map $z\mapsto z^2$. From this we infer that there exists a non empty open set $U\subset \hat{\mathbb{C}} \backslash h\circ g(D)$.
Why do we know that such a $U$ exists?
Why could we not find an open set $U' \subset \hat{\mathbb{C}} \backslash g(D)$?
There are simply connected domains that are dense in the sphere, $\mathbb{C}\setminus \{ t\in \mathbb{R} : t \leqslant 0\}$ for example. Since Möbius transformations are automorphisms of the sphere, $D$ is dense if and only $g(D)$ is dense. If $D$ is not dense, we don't need to apply $h$, since then (by definition) there already is a nonempty open set contained in $\hat{\mathbb{C}} \setminus g(D)$, but applying $h$ doesn't harm, so we avoid the case distinction by always applying it. We can guarantee that there is always a nonempty open set contained in $\hat{\mathbb{C}} \setminus h(g(D))$ because every nonzero complex number has exactly two square roots, of which $h$ chooses one, so $w \in h(g(D)) \implies -w \notin h(g(D))$, and if $W\subset h(g(D))$ is a nonempty open set, then $-W$ is a nonempty open set contained in the complement of $h(g(D))$.