I have to prove there is no smooth map $f:\mathbb{R}^3 \to \mathbb{R}$ and no regular value $y$ of $f$ such that $f^{-1}(y)$ is the projective space of dimension 2.
From the pre-image theorem, $f^{-1}(y)$ is either empty or a closed sub manifold of cxdimension 1, but I don't know how to conclude it's not the projective plane.
As an application of Jordan-Brower separation theorem, one can prove that no non-orientable hypersurface can be embedded in $\mathbb{R}^n$. Thus, $\mathbb{R}P^2$ cannot be embedded in $\mathbb{R}^{3}$.