By 'open map' I mean that the quotient map $q\colon S^2 \to \mathbb{RP^2}$ maps open sets in $S^2$ to open sets in the projective plane $\mathbb{RP^2}$.
My thinking is the following:
$q\colon S^2 \to \mathbb{RP^2}$ is an open map $\iff \forall$ open $U \subseteq S^2,\ q(U) \subseteq \mathbb{RP^2}$ is open $\iff \forall$ open $U \subseteq S^2,\ q^{-1}(q(U)) \subseteq S^2$ is open. Now, $q^{-1}(q(U)) = U$ so we must have that $q$ is an open map because $U$ is open in $S^2$.
Is this correct? Any feedback would be appreciated!
Almost. We have $q^{-1}(q(U))=U\cup(-U)$. And that is open as union of open sets.