I’m missing a step in Lang’s proof of the Holomorphic Inverse Function Theorem I’m looking at the proof of theorem 1.7 in chapter VI of Lang’s Complex Analysis (page 182 of the 4th edition of the book), and I’m stuck with one of the last statements. More precisely:

The argument we have given also shows that $f$ is an open mapping.

Forgive me for not writing a concise sketch of the steps of the proof. I’d say it’s better to read it from the start (it is pretty short).
My problem is that it seems to me that the given argument only shows that open sets of a certain kind have open image, namely those of the form $\left\{ |z|<r, |f(z)|< \frac{r}{2} \right\}$ for suitably small $r$‘s, and I don’t see how to deduce from that that all the other open sets have open image as well.
Any help is greatly appreciated.
 A: The argument says:
(1) there exists a bijection $f|_{U}:U\to D(0,\frac{r}{2})$;
(2) the set $U$ is an open set containing $0$ that can be arbitraily small (in the sense it is contained in a ball with arbitrarily small radius, and hence we can assume $U\subset\mathcal O$).
By a translation you can generalize this to (for any open set $\mathcal O$):
(1) for every $z\in\mathcal O$, there is a bijection $f|_{U_z}:U_z\to D(f(z),\frac{r_z}{2})$;
(2) each $U_z$ is an open set containing $z$ that can be arbitraily small (in the sense it is contained in a ball with arbitrarily small radius, and hence we can assume $U_z\subset\mathcal O$).
And being locally open is enough, since you can write
$$\mathcal O=\bigcup_{z\in\mathcal O}U_z$$
where $U_z$ is chosen as above. Then
$$f(\mathcal O)=\bigcup_{z\in\mathcal O}f(U_z)=\bigcup_{z\in\mathcal O}D(f(z),\frac{r_z}{2})$$
Recall that any union of open sets is open, from which it follows the open mapping theorem of analytic functions, which is used to prove the inverse mapping theorem.
P.S. I would suggest you to read Conway's proof on this. The method is slightly different, and maybe you can benefit from comparing these two.
