Suppose $p:S^2\to P^2$ is the quotient antipodal map, and $J$ is a simply closed curve in $P^2$, then $p^{-1}(J)$ is either a simply closed curve in $S^2$, or two disjoint simply closed curves in $S^2$.
I think I can prove the second case. Clearly $(S^2,p)$ is a 2-fold covering space. Suppose $J$ is totally contained in a basic neighbourhood $U$, of $P^2$, then $p$ is a homeomorphic projection on the two connected components of $p^{-1}(U)$, hence each component contains exactly one homeomorphic preimage of $J$.
Now I have trouble proving the other case where no basic neighbourhood can contain $J$. How to proceed? Thanks in advance.
$S^2 \to P^2$ is a 2-sheeted covering map. Now note, that for any space $X\subset P^2$, the map restricts to a 2-sheeted covering map $p^{-1}X\to X$. If we pick $X=J$, then the covering map restricts to
$$ p^{-1}J \stackrel {2:1} \longrightarrow J\cong S^1.$$
So the preimage $p^{-1}J$ might either be homeomorphic to the unique covering space of $S^1$ corresponding to $2\mathbb Z\subset \mathbb Z$, or be homeomorphic to the trivial covering space $S^1 \sqcup S^1$. Note that generally (not in this case for the Jordan curve theorem) the two curves might be knotted.