Nowhere vanishing vector field on Moebius strip I know there is no continuos non-vanishing normal vector field on Moebius strip, which is pretty obvious. Is it possible to construct a nowhere vanishing tangent vector field on Moebius strip? Intuitively it looks possible. Make it parallallel to the x-axis of the tangent plane (parallel to the middle circle). When the orientation flips, we only flip the y-axis, so our vector field stays continuous. I can't fully grasp this.. If this works, then how to rigorously prove continuity?
 A: Proving this rigorously is as simple as giving a rigorous definition of the Mobius strip. Here's one:
Take $\Bbb R \times (-1,1)$ and identify $(x,t)$ with $(x+1, -t)$ for all $(x,t)$. The quotient, $M$, is the same thing you get if you took $[0,1] \times (-1,1)$ and identified $(0,t) = (1,-t)$.
Your vector field upstairs on $\Bbb R \times (-1,1)$ is the right idea: let's make it parallel to the forward-direction on the Mobius strip (as opposed to the direction that takes us off the edge). Write $V(x,t) = \langle 1, 0\rangle$, pointing right at every point (thinking of $(-1,1)$ as a vertical factor). 
When we do the translation-and-reflection $(x,t) \mapsto (x+1,-t)$, the vector field first gets translated (nothing changes, since it's translation-invariant), and then reflected across the $x$-axis (so $V(x,t) = \langle a, b\rangle$ would be sent to $\langle a, -b\rangle$); but since our vector field is $\langle 1, 0\rangle$, it's kept fixed by this flipping. 
If this is confusing, try visualizing it instead on $[0,1] \times (-1,1)$: draw the vector field, and see what happens to it when you glue the left side to the right side with a twist. (Bonus points: picture why this doesn't work for the vector field $\langle 0, 1\rangle$.)
Anyway, $V(x,t)$ descends downstairs to $M$, and gives your desired vector field. 
