Of course there are many ways to prove this. However, I came across the following exercise (Ch. 0 #3).
Prove that: (a) a regular surface $S\subset \mathbb{R}^3$ is an orientable manifold if and only if there exists a differentiable mapping of $N:S\rightarrow \mathbb{R}^3$ with $N(p)\perp T_p(S)$ and $|N(p)|=1$, for all $p\in S$. (b) the Möbius band (Example 4.9 (b)) is non-orientable.
In Example 4.9 (b), he constructs the Möbius band as the quotient by the antipodal map of the cylinder $C=\{(x,y,z)\in \mathbb{R}^3:x^2+y^2=1,|z|<1\}$. The problem, of course, is that this isn't given as a surface in $\mathbb{R}^3$! I was thinking for a second that maybe I should try and construct a map $C\rightarrow S^2$ with the right properties and check that it doesn't descend to a map on $M$, but that's stupid because if I were to embed $M\subset \mathbb{R}^3$, I'm pretty sure it couldn't possibly have those tangent planes anyways.
Does anyone have any insight? Presumably the solution to (b) should use the fact given in (a).