A Covering Map $\mathbb{R}P^2\longrightarrow X$ is a homeomorphism I came across the following problem:  Any covering map $\mathbb{R}P^2\longrightarrow X$ is a homeomorphism.  To solve the problem you can look at the composition of covering maps 
$$
S^2\longrightarrow \mathbb{R}P^2\longrightarrow X
$$
and examine the deck transformations to show that the covering $S^2\longrightarrow X$ only has the identity and antipodal maps as deck transformations.
I've seen these types of problems solved by showing that the covering is one-sheeted.  Is there a solution to the problem along those lines?
EDIT:  Even if there isn't a way to do it by showing it is one-sheeted, are there other ways?

1-Prove that $X$ has to be a compact topological surface;
2-Prove that such a covering has to be finite-sheeted;
3-Deduce from 2 and from $\pi_1(\mathbb{R}P^2)$ that $\pi_1(X)$ is finite;
4- Since the map induced by the covering projection on $\pi_1$ is injective you get $\mathbb{Z}/2\mathbb{Z}< \pi_1(X)$;
5-Conclude using the classification of compact topological surfaces.
