Take two Moebius bands glued together as in the picture here. Now attach a $2$-cell along one of the boundary components of this space (indicated in green in the picture). Call the resulting space $X$. My gut feeling is that $X$ cannot be embedded in $\mathbb{R}^3$. The question is: is this actually true, and if so, how do I prove this?
I thought of approaching this problem via Alexander Duality (which basically tells you that the homology groups $H_i$ of a compact and locally contractible subset of $\mathbb{R}^n$ are trivial for $i \geq n$ and torsionfree for $i=n-1$ and $n-2$), but $H_i(X)$ are -- I think -- trivial for $i \geq 2$ and $\mathbb{Z}$ for $i=0$, $1$.
I also tried embedding a complete graph on six vertices into $X$ minus the $2$-cell and then arguing as in Maehara's paper "Why is $P^2$ not embeddable in $\mathbb{R}^3$?" (see also Gjergji Zaimi's answer here), but I can't seem to get it right.
