This is exercise III.4.15 of Freitag and Busam.
Exercise: Let $g$ be an analytic function on an open set containing the closed disk $\bar{U}_r(a)$. We assume $|g(a)|<|g(z)| $ for all $z$ on the boundary of the disk. Then there exists a zero of $g$ in the interior of the disk.
Using this, find a further proof of the Open Mapping Theorem.
Solution: If $f$ is a non-constant function on a domain $D\supset \bar{U}_r(a)$, then there exists an $\varepsilon>0$ with $|f(z)-f(a)|\geq 2\varepsilon$ for $|z-a|=r$. (*)
Then, using the first part of the exercise it is easy to see that $U_\varepsilon (f(a))$ is contained in the image of $f$. (**)
I can prove the first part using the maximum principle, but I don't understand why (* ) is true nor why (* ) and the first part imply (**).
Thanks in advance for your help!