I'm looking for verification of my proof of the following statement:
Let $f \colon \mathbb{D} \to \mathbb{C}$ be an injective and holomorphic function on the unit disc which satisfies $f(0) = 0$ and $f'(0) = 1$. Show that $$\inf\{\vert w \vert : w \not\in f(\mathbb{D})\} \leq 1$$ with equality if and only if $f(z) = z$ for all $z \in \mathbb{D}$.
The given values of $f$ and $f'$ are hypotheses of the Schwarz lemma, so I tried to use that. If the claimed inequality is false or we have equality, then $f(\mathbb{D})$ contains $\mathbb{D}$, so we may set $U = f^{-1}(\mathbb{D})$. Since $f$ is injective, $f$ has an inverse $g = f^{-1}$ on $U$. This gives us a holomorphic function $g \colon \mathbb{D} \to U$ whose image $U$ lies in $\mathbb{D}$. Moreover, since $f(0) = 0$ and $f'(0) = 1$, we have $$g(0) = 0, g'(0) = \frac{1}{f'(g(0))} = 1.$$ Therefore the Schwarz lemma says that $g(z) = \alpha z$ with $\vert \alpha \vert = 1$. By analytic continuation, the inverse of $f$ on $f(\mathbb{D})$ must also be $\alpha z$, thus $f(z) = \alpha^{-1}z$, and $f'(z) = \alpha^{-1}$, which forces $\alpha = 1$. Equality is clearly satisfied for $f(z) = z$.