Moebius band not homeomorphic to Cylinder. I have been trying to think of a rather basic way of proving this, but it seems a bit elusive. In the case with boundary (looking at them as quotients in $ [0,1] \times [0,1]  $), they can be distinguished from the connectedness of the boundary (thanks Stefan), but I'm interested in the open case (looking at them as quotients in $ [0,1] \times (0,1)  $, i.e. as manifolds without boundary).

If there is not such basic way to do this, it would be interesting to read about some advanced methods you guys know about.
Here is the most elementary argument I can come up with.  Let $C$ denote the open cylinder and $M$ denote the open Moebius band.  Note that there exists a compact subset $K\subset C$ such that for any compact $K'$ such that $K\subseteq K'\subset C$, $C\setminus K'$ is disconnected (for instance, you can take $K$ to be the "equator" of $C$).  On the other hand, I claim that no compact subset $K\subset M$ has this property.  Indeed, let $K\subset M$ be compact; identifying $M$ as a quotient of $[0,1]\times(0,1)$ in the standard way, there is then some $\epsilon&gt;0$ such that $K$ is contained in the image of $[0,1]\times[\epsilon,1-\epsilon]$ under the quotient map (for instance, because the images of sets of the form $[0, 1] \times (\epsilon, 1-\epsilon)$ are open and cover $K$). Let $K'$ be this image; then $K'$ is compact, $K\subseteq K'\subset M$, and $M\setminus K'$ is connected.
Let's be clear we have $X = [0,1]\times (0,1)$ with two different identification maps:


$f:(0,x) \in X  \mapsto (1,x) \in X$ this gives a cylinder $Y$
$f:(0,x) \in X  \mapsto (1,1-x) \in X$ this gives a Möbius band $Z$


Even though $Y$ and $Z$ are open, we can take their closure and $\partial \overline{Y} = S^1 \cup S^1$ while $\partial \overline{Z} = S^1 $.  See Chapter 4 of the notes of Allen Hatcher on Point Set Topology for a discussion on quotient spaces.



#1 Consider the subspace $X_0= (0,1) \times [\frac{1}{4}, \frac{3}{4}]$ common to both spaces.  Let $f:Y\to Z$ be a homeomorphism and the closure $\overline{X_0}\subset Y$ is a cylinder while $\overline{f(X_0)} \subset Z$ is a Mobius band.  



#2 Cylinder is orientable.  Mobius band isn't.



#3 Remove the meridian circle $ [0,1] \times \frac{1}{2} $ from a cylinder - it is not connected.  Remove a meridian circle from a Möbius band, it remains connected.  Jordan curve is not necessary, merely that if a continuous map $\phi:[0,1] \to \mathbb{R}$ has positive and negative values, then $\phi(x) = 0$ somewhere.