= Solution
A <topological surface> is a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of $\mathbb R^2$.
The antipodal action on $S^2$ is free. The <surface quotient by a free finite action> applies: Around each point choose a small <open disc> disjoint from its antipodal image; the quotient map restricts to a homeomorphism from that disc onto an open neighbourhood in the quotient. Compactness gives Hausdorffness and second countability descends from the sphere. Thus the quotient is the <real projective plane>, in particular a topological surface.
For the second quotient write a point away from the poles as
$$
(\sqrt{1-z^2}\cos\phi,\sqrt{1-z^2}\sin\phi,z).
$$
The map
$$
q(\phi,z)=(2\phi,z)
$$
in these cylindrical coordinates extends continuously over the poles and identifies exactly $\phi$ with $\phi+\pi$. It therefore induces a continuous bijection $S^2/\!\approx\,\to S^2$. The domain is compact and the sphere Hausdorff, so this bijection is a homeomorphism.
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