It was taken from Complex Analysis (Princeton Lectures in Analysis, No. 2) by Elias M. Stein and Rami Shakarchi page 92.
This is the rough idea of the proof of open mapping theorem:
Theorem: If $f$ is non-constant holomorphic function in a region $\Omega$, then $f$ is open.
Let $w_0$ belong to the image of $f$, say $w_0=f(z_0)$. We must prove that all points $w$ near $w_0$ also belongs to the image of f. Define $g(z) = f(z) - w$. Choose $\delta \gt 0 $ s.t. the disc $|z-z_0| \lt \delta$ is contained the the domain $\Omega$. Then at the end it proves that $g$ has a zero inside the circle $|z-z_0| = \delta$.
Why is it the end of the proof? Can someone elaborate a little bit more?
And this is the rough idea of the proof of maximum modulus principle:
Theorem: If $f$ is non-constant holomorphic function in a region $\Omega$, then $f$ cannot attain a maximum in $\Omega$.
Suppose that $f$ did attain a maximum at $z_0$. Since $f$ is holomorphic it is an open mapping, and therefore, if $D \subset \Omega $ is a small disc centered at $z_0$, its image $f(D)$ is open and contains $f(z_0)$. This proves that there are points in $z \in D$ such that $|f(z)| \gt |f(z_0)|$, a contradiction.
How comes there are points in $z \in D$ such that $|f(z)| \gt |f(z_0)|$? Did I miss some basic stuffs?