Let $M$ a surface, $\dim M=m$, and let $\omega$ a continuous $m-$form in $M$. I want to prove that $M$ is orientable $\iff \omega(x)\neq 0\; \forall x\in M$.
The $(\Leftarrow)$ part is ok to me, since I say this post. However, I don't know exactly how to prove the $(\Rightarrow) $ part, even if it looks kind of obvious.
I tried to taking two parametrizations $\phi,\psi$ of $x\in M$ in an orientation of $M$ and work with writting $\omega$ in each basis, but I don't know exactly how to proceed. Any tips?
The statement as written is false. On any manifold (orientable or otherwise), there exists a continuous $m$-form that is equal to zero everywhere. That is a trivial counterexample.
I think the statement should say, if $M$ is orientable then there exists a continuous $m$-form $\omega$ such that $\omega(x) \neq 0$ for all $x \in M$.
To construct this $m$-form, I suggest you take a positively-oriented atlas, and construct positive-valued $m$-forms locally in each coordinate chart. Then multiply these local $m$-forms by a partition of unity (so that they can be extended continuously to the entire manifold), and add them together. Use the fact that the Jacobian determinants are positive (since the charts are positively-oriented with respect to each other) to conclude that the sum of the local $m$-forms is also positive in each chart.