I have some understanding problems with the open mapping theorem.
Lemma
Let $f\in H(D(z_{0},\ R))$ ($f$ analytical on the disc with the origin in $z_0$ and radius $R$) and $ |f(z_{0})|<\min_{|z-z_{0}|=r}|f(z)| $ for a $r\in(0,\ R)$. Then there is a $w\in D(z_{0},\ r)$ with $f(w)=0$.
Theorem
Let $\Omega\subseteq \mathbb{C}$ be a region and $f\in H(\Omega)$. Then we have either $f(\Omega)$ is also a region or $f$ is constant.
Proof
We choose $f$ not constant. $f(\Omega)$ is connected as a continuous image of a connected set. We show, that $f(\Omega)$ is open:
Let $w_{0}\in f(\Omega)$, so $w_{0}=f(z_{0})$ for a $ z_{0}\in\Omega$. There is a $\delta>0$ with $ f(z)\neq w_{0}\ (z\in\partial D(z_{0},\delta)),\ (a) $ because otherwise $f(z)=w_{0}$ for all $ z\in\Omega$.
$\color{green}{\text{I don't understand why there is a } \delta \text{ so that (a) is fulfilled. I don't understand the followup.}}$
$\color{green}{\text{What is actually the outline of proving the openness in the next steps?}}$ $\color{green}{\text{(Read: I have no idea what happens in the steps below)?}}$
Because $f$ is continuous, there is a $\varepsilon>0$ with $ |f(z)-w_{0}|\geq 3\varepsilon\ (z\in\partial D(z_{0},\delta)). $ We show: $D(w_{0},e)\subseteq f(\Omega)$: For $|w-w_{0}|<\varepsilon$ and $|z-z_{0}|=\delta$ we have: $ |f(z)-w|=|f(z)-w_{0}+w_{0}-w|\geq|f(z)-w_{0}|-|w_{0}-w|\geq 3e-e=2e $ and $|f(z_{0})-w|=|w_{0}-w|<\varepsilon$, so that $ |f(z_{0})-w|<\min_{|z-z_{0}|=\delta}|f(z)-w|. $ Using the Lemma presented before the theorem, we have that $f(z)-w$ has a zero $z_{w}$ with $|z_{w}-z_{0}|<\delta$, so that $f(z_{w})=w$ with $ z_{w}\in\Omega.\ \square $