I'm studyng De Rham theorem for smooth manifolds and I noticed a thing. Many authors prove the theorem for a convex subset of $\mathbb{R}^n$ and then they show that the theorem is true for any open subset of $\mathbb{R}^n$. Defined a De Rham basis for the manifold $M$ as a basis $\mathcal{B}$ such that the theorem it's valid for each $U\in\mathcal{B}$ they check that each manifold with De Rham basis is De Rham, i.e. it's valid the De Rham Theorem. At this point the proof is finished since by definition a smooth manifold has a basis of open subset diffeomorphic to open subset of $\mathbb{R}^n$. Now my question is the following...According with "Introduction to Smooth Manifolds" by J.M. Lee every smoth manifold has a basis of subsets diffeomorphic to $\mathbb{R}^n$ and this basis is obvsliuly a De Rham basis, then what is utility to prove the De Rham Theorem valid for any open subset of $\mathbb{R}^n$?
The approach described above is used, for example, in the same book of J.M. Lee.
Why do you think that a basis of subsets diffeomorphic to $\mathbb R^n$ is "obviously" a de Rham basis? To be a de Rham basis means that each basis set and all finite intersections of basis sets satisfy the de Rham theorem. In general, a finite intersection of subsets diffeomorphic to $\mathbb R^n$ is diffeomorphic to an open subset of $\mathbb R^n$, but you can't say much more about it than that. That's why it's necessary to prove first that the de Rham theorem holds for such subsets.