This "proof" of mine sounds deceivingly simple, and I think the result is too general to hold: also, I can't find any confirmation online. I would be grateful is somebody could have a look, and maybe find something suspicious.
So, I'm given a section of a fibration, and I want to check whether it's an open map. I know $\pi\cdot s(U)=U$, for $U$ a set in the base space; then $\pi^{-1}\cdot\pi\cdot s(U)=\pi^{-1}(U)$, and by surjectivity of $\pi$ I get $s(U)=\pi^{-1}(U)$. But $\pi$ is continuous, thus $s$ must be open.
I'm not sure that in a fibration $\pi$ is always taken to be surjective - but it is in the case I am considering, anyway.
Comments? Remarks? Am I missing something easy?
$\pi$ is surjective iff for all subsets $B$ of the image $\pi[\pi^{-1}[B]] = B$, so I'm not quite sure how you use the "from surjectivity of $\pi$" part to get to the final formula. You seem to use $\pi^{-1}[\pi[A]] = A$ for subsets of the domain, but this is equivalent to injectivity.