As with most questions where we want to keep from getting into deep arguements, it is a questions of how deep can we go? Here is what I came up with for a intuitive arguement:

First, we notice that we can embed (Put in so it doesn't really change) the cylinder $C$ into $\mathbb{R}^2$. Just think of sitting the cylinder on the table, putting your hands inside and pushing the sides down and out away from the center.

If the cylinder and Möbius strip were homeomoprhic, we could also embed it in the plane. But we cannot, and here is why:

Any map from the Möbius strip into the plane will not be injective, which means that (at least) two points will be sent to the same place, no matter what. Here is a Möbius strip, with the two sides identified.

And here is the same Möbius strip with two circles that *only intersect once* in the Möbius strip. (This is the important part.)

If we wanted to embed the Möbius strip, we would also embed the circles. So lets put the red circle in first, standardly, and the try to put in the blue circle.

We can see that blue circle *should* be on both sides of the red circle, but for that to happen, they *must* intersect each other twice! So, we can't embed the Möbius strip into the plane, hence, the cylinder and the Möbius strip are not homeomorphic.