 Solution

The map $f$ is bounded by one near the puncture, so the Riemann removable singularity theorem extends it holomorphically to a map $\tilde{f}:\Delta \to \mathbb{C}$. The maximum modulus principle shows that its image still lies in $\Delta$: if $|\tilde{f}(0)|=1$, the extension would be constant, contradicting its values in $\Delta \setminus \{0\}$ unless that constant also had modulus less than one.

If $f$ is biholomorphic, its inverse $g:\Delta \setminus \{0\}\to \Delta \setminus \{0\}$ also extends holomorphically. Were $a=\tilde{f}(0)\ne 0$, continuity of $\tilde{g}\circ \tilde{f}$ and the identity $g(f(z))=z$ on the punctured disk would give $g(a)=0$. This is impossible because $a\in \Delta \setminus \{0\}$ and $g$ takes values in the punctured disk. Hence