Consider the Moebius strip $M$ as a quotient space of $\mathbb{R}\times [0,1]$ with opposite lines glued together with reverse orientation. I would like to prove that a closed submanifold $N$ in $\mathbb{R}^3$ cannot be diffeomorphic to $M$.
It seems to me that, even if $M$ is not the classical Moebius band with its boundaries, $M$ is still a non-orientable manifold.
My plan is to prove that any closed 2-submanifold $N$ of $\mathbb{R}^3$ is orientable and thus may not be diffeomorphic to $M$.
It seems to me that answering the two following questions is a good way to prove it:
1) Show that any closed submanifold of dimension 2 in $\mathbb{R}^3$ can be seen as $f^{-1}(0)$ where $f : \mathbb{R}^3 \rightarrow \mathbb{R}$ is a smooth map such that $0$ is a regular value of $f$.
2) Show that any closed submanifold of dimension $2$ of $\mathbb{R}^3$ that is obtained as : $f^{-1}(0)$ where $f : \mathbb{R}^3 \rightarrow \mathbb{R}$ is a smooth map, such that $0$ is a regular value of $f$, is orientable.
Any idea on how to tackle 1) and 2)?
I'm not sure I fully understand what your $M$ is. It does not look compact to me. Anyway it is well-known that a closed 2-submanifold of $\Bbb R^3$ must be orientable. This is a consequence of the Alexander Duality Theorem. See Corollary 3.45 of Hatcher's Algebraic Topology.