Can you cut a Mobius band to produce two Mobius bands? I had a homework problem that asked "Can you cut a Klein bottle into two Mobius bands?" which I carelessly read initially as "Can you cut a Mobius band into two Mobius bands? After realizing my mistake, I quickly demonstrated that the actual problem of cutting a Klein bottle into two Mobius bands, but after spending an hour or so attempting to cut a Mobius band into two Mobius bands I am curious if it is possible... My intuition is no, but I am unable to prove this rigorously. The best I could do was cutting a Mobius band into a Mobius band and a cylinder with two twists (still a cylinder though). 
So does there exists a cutset $C$ of the Mobius band $M$ such that $C$ is a simple closed curve in $M$ and $M \smallsetminus C$ is the union of two disjoint open sets such that the closure of each in $M$ is homeomorphic to $M$?

First a small point of clarification, I'm not sure if the original asker was thinking about this, but a simple closed curve is generally assumed to lie in the interior of a manifold with boundary (otherwise cutting along it wouldn't make much sense!).
With that in mind, there's a nice quick way to see that this is impossible. Suppose you have a simple closed curve cutting your Möbius band into two pieces. Both of those two pieces must have a boundary component corresponding to the curve you cut along. So, if we want those pieces to be Möbius bands, we have to be able to glue them back up along that boundary component to get a Möbius band. But the result of that operation has no boundary as we glued along all of the boundary components, so it certainly can't be a Möbius band! (and as you note is in fact a Klein bottle).
A more systematic way might be to list out the isotopy classes of simple closed curves on the Möbius band (there's only 2 of them) and check the result of cutting along a representative of each.
