 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 $\alpha$ 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 $\alpha$ 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.