My approach was to the following:
Let G be a domain, because of the Riemann mapping theorem $G$ is equivalent to the unit disc $\mathbb{E}$. With $z_0 \in G$ follows that $G\setminus \{z_0\}$ is equivalent to $\mathbb{E}\setminus\{0\}$. Since $z_0$ was arbitrary $\bigcap_\limits{z_0\in[-1,1]}G\setminus \{z_0\}$ is equivalent to $\mathbb{E}\setminus\{0\}$.
I am now thinking about 'expanding' $G$ to $\mathbb{C}$ but I'm not allowed to set $G=\mathbb{C}$ because then I can't use the Rieman mapping theorem. Is everything I did correct so far or is there another way of proving this?
Consider first $U=\Bbb C_\infty\setminus[-1,1]$. By a Mobius transformation we can map $U$ to the plane $V$ cut along the negative real axis. Taking the square root and then another Mobius transformation takes $V$ to $D$, the unit disc.
If we delete a point from $U$ we find that $U\setminus\{a\}$ is conformally equivalent to $D\setminus\{b\}$ for some $b$. We can take $a=\infty$. Whatever $b$ is, a further Mobius transformation takes $D$ to itself, and $b$ to $0$.