2-dimensional manifold is orientable if and only if it does not contain the Möbius band

Question

Criterion for non-orientability: a manifold $M$ is non-orientable if and only if it contains a "bad" sequence of charts, i.e. $\{U_i,\varphi_i\},i=1,\ldots,n$ such that the transition maps $\varphi_{i+1}\circ\varphi_i^{-1},i=1,\ldots,n-1$ all have positive Jacobian determinant, and $\varphi_1\circ\varphi_n^{-1}$ has negative Jacobian determinant at some point $x\in\varphi_1(U_1\cap U_n)$. The Möbius band is non-orientable. So, one part of the question is clear: given that $M$ is orientable, suppose that it contains the Möbius band. Then in the Möbius band there's a "bad" sequence of charts. Then it's also such a sequence in $M$, so it is non-orientable; a contradiction. My question is about the converse. Given that there's no Möbius band within $M$, suppose $M$ is non-orientable. In order to arrive at a contradiction, we'd have to construct a Möbius band in $M$. How do I approach this? Any help would be appreciated.