My question is as follows:
Let G be a finite group which acts freely as a group of homeomorphisms of a closed surface S (so the only element with fixed points is the identity)
Then: Show the orbit space $S/G$ is a closed surface Show S may be orientable while S/G isn't If S/G is orientable, must S be?
Help with the first out of these would be
On
Since the group acts freely, for any point $x \in S$, the elements $g \cdot x$ are pairwise distinct for $g \in G$. Because $S$ is Hausdorff, this means that for every $g$, there are neighborhoods $U_g$ and $V_g$ respectively of $x$ and $g \cdot x$ that are disjoint. Now $G$ is finite so $\bigcap_{g \in G} U_g$ is still a neighborhood of $x$. $G$ acts by homeomorphism, so if you restrict the neighborhoods you can then prove that the quotient map is a local homeomorphism. Thus $S/G$ is locally homeomorphic to a plane, just like $S$.
PS: If you know what "properly discontinuous action" means, I just told you that $G$ acts properly discontinuously on $S$.
Similarly if $x$ and $y$ have different orbits, then $g \cdot x \neq h \cdot y$ for all $g,h \in G$ and so you can find neighborhoods $U_g$, $V_h$ that don't intersect. The intersections $\bigcap_g U_g$ and $\bigcap_h V_h$ are still neighborhoods because $G$ is finite, and map to neighborhoods of $[x]$ and $[y]$ that don't intersect in $S/G$. So $S/G$ is Hausdorff too.
For the second question: Let $S$ be the 2-sphere $S^2$, and $G = \mathbb{Z}/2\mathbb{Z}$ act by the antipodal action to get $\mathbb{RP}^2$.
Hint: Show that the quotient map $S \to S/G$ is a finite-sheeted covering to deduce that $S/G$ is a closed surface. Also, an interesting example to consider is the action $\mathbb{Z}_2 \curvearrowright \mathbb{S}^2$ whose quotient is $\mathbb{R} \mathbb{P}^2$.