= Solution
Near an isolated point $P\in A$, choose source and target coordinate charts, using continuity of $\alpha$ to keep a sufficiently small source neighbourhood in the target chart. The coordinate representation is holomorphic on the punctured neighbourhood and is bounded near $P$ by continuity. The <Riemann removable singularity theorem> extends it holomorphically across $P$, with the given continuous value. Repeating at all points of $A$ proves that \b[$\alpha$ is analytic].
For a nonconstant holomorphic function $f$ on a connected <Riemann surface>, let $A$ be the set where its coordinate derivative vanishes. This condition is independent of the chart. The derivative cannot vanish identically on an open set, because the <identity theorem> would then make $f$ constant globally. Its zeros are isolated and closed locally, so $A$ is a closed discrete subset. At every other point the <holomorphic inverse function theorem> makes $f$ a local coordinate chart.
Now suppose $\alpha$ is a homeomorphism and $g=f\circ\alpha$ is analytic. Outside $\alpha^{-1}(A)$, use $f$ as a target coordinate to write locally $\alpha=f^{-1}\circ g$. This is analytic. The preimage $\alpha^{-1}(A)$ is discrete because $\alpha$ is a homeomorphism, and the first argument extends analyticity across it. An injective holomorphic map cannot have a zero derivative: a zero of order at least two would give a locally multiple-to-one map. Its local inverse is therefore holomorphic, proving that \b[$\alpha$ is a conformal equivalence].
The plane $\mathbb C$ and the unit disk are homeomorphic by $z\mapsto z/(1+|z|)$, with inverse $w\mapsto w/(1-|w|)$. They are not conformally equivalent: a holomorphic equivalence from the plane to the disk would be a bounded nonconstant <entire function>, contradicting <Liouville's theorem>.
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