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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.

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    The Mayer-Vietoris sequence and the five Lemma are a powerful tool for proving inductively that cohomology or homology groups for different theories are equal. It's more or less by induction. – 2017-02-26

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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 $\mathbb{R}^n$?

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.

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    First of all...it's an honor for me to speak with you Professor Lee, I love your books :) Now, this is my idea: Every point of the manifold $M$ has a neightborhood diffeomorphic to $\mathbb{R}^n$ and therefore I can take a basis of $M$ formed by open sets diffeomorphic to $\mathbb{R}^n$. If I consider their finite intersection then I have a set diffeomorphic to an open and convex subset of $\mathbb{R}^n$, but any starshaped open subset of $\mathbb{R}^n$ is diffeomorphic to $\mathbb{R}^n$ and therefore this claim holds in particular for convex subsets. Is this idea correct? – 2017-03-04
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    Thanks for the kind compliment! The problem with this argument is that there's no reason the intersection should be convex, or diffeomorphic to $\mathbb R^n$. Think of the two charts on $\mathbb S^2$ obtained from stereographic projection from the north and south poles. – 2017-03-04
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    Thanks for the helpfull remark Professor Lee :) Now I understand my mistake! – 2017-03-04