Q:

Open Mapping Theorem proof

In my notes on complex analysis we have this version of the Open Mapping Theorem

The image of an open set under a non-constant holomorphic map is open.

The proof begins by stating that it is enough to show that the image of an open disc is open under a non-constant holomorphic map.
I do not understand why this is the case.

A:

Every open set is an union of open discs. If you prove that the image of an open disc is open, then, for every open set $U$ we have
$$U=\bigcup_{i\in I}D_i$$
$$f(U)=f\left(\bigcup_{i\in I}D_i\right)=\bigcup_{i\in I}f(D_i)$$
So if each $f(D_i)$ is open, then $f(U)$ is open.

