# Proving that a closed, non-orientable surface cannot be embedded in $\mathbb{R}^{3}$ using Euler characteristic

So, though I am aware that it is conventional to use Alexander's duality theorem to prove that no closed, non-orientable surface $$S$$ can be embedded into $$\mathbb{R}^{3}$$; I was hoping perhaps to find a more elementary version of this proof by the consideration of the Euler characteristic of $$S$$ (as is done for the answer to this question along with Alexander's duality). I wish to prove this theorem for the sake of proving the inscribed rectangle conjecture.

The argument I am considering goes like this (with reference in particular to the closed surface formed by pasting the boundary of a disk $$D$$ to that of a Mobius loop $$M$$, whose boundary is mapped to a jordan curve. We assume that this mapping is a homeomorphism, in order to cause a contradiction): To begin with, as $$\chi(M) = 0$$ and $$\chi(D) = 1$$. Then, we know that $$\chi(M \cup D) = \chi(M) + \chi(D) -\chi(M \cap D)$$ As the boundary of a Disk (i.e. the space $$M \cap D$$) is a loop (with Euler characteristic $$0$$),
$$\chi(M \cup D) = \chi(M) + \chi(D) -\chi(M \cap D) = 0 +1-0 =1$$ Which is odd.
It is here that I get stuck, as I do not know how to use this to show somehow that this leads to a contradiction to our assumption that the function mapping the Mobius loop into $$\mathbb{R}^{3}$$ (where the boundary is a jordan curve) leads to a contradiction.
Is there any theorem that says, for example, that any closed surface in $$\mathbb{R}^{3}$$ must have an Euler characteristic that is an even number?

If this proof is insufficient, then how else may I prove it in a manner that is at-least semi rigourous whilst still being accessible to an undergraduate who has just started to learn topology?