I know there are already many posts about this question, and I've read and understood the proof in Stein and Shakarchi, but I don't understand why my “shortcut” is wrong.
Assume $f$ is injective and holomophic in $\Omega$ open set and $f'(z_0) = 0$ for a $z_0 \in \Omega$.
Consider $g(z) = f(z_0+z)-f(z_0)$, this way $g(0) = g'(0) = 0$ and $g$ is holomorphic and injective in $\Omega-z_0$.
Now, since $g$ and $g'$ are analytic and non-constant, $0$ is an isolated zero for both of them, that is, there exists a disc $D \subseteq \Omega-z_0$ around $0$ such that $g$ and $g'$ have no zeroes in $D \setminus \{0\}$.
Take $w$ such that $0 < |w| < \min_{\partial D} |g|$, then for Rouché's Theorem $g$ and $g-w$ have the same number of zeroes (counting the multiplicity) in $\mathring{D}$.
Since $0$ is a zero of multiplicity at least $2$ for $g$, $g-w = 0$ has at least $2$ solutions in $\mathring{D} \setminus \{0\}$ which are different, because otherwise $g'$ would have to be $0$ in that point, leading to a contradiction.
