How does one prove that the Klein bottle cannot be embedded in $R^3$?
Any compact, smooth hypersurface in $\Bbb R^n$ is necessarily orientable. (Without fancy algebraic topology, this is a consequence of the Jordan-Brouwer Separation Theorem.) Indeed, more is true: A compact hypersurface in any simply connected manifold is orientable.
MO answer. In the link, there is a reference to a proof in Hatcher using Alexander duality as well as a more elementary proof.