requirement of openess of the subset of a manifold for mayer-vietoris theorem So for a given manifold $M$ and two of $U,V$ open sets that can be used to cover $M$, one can use to mayer-vietoris theorem to relate the decomposed de rham cohomology of $M$ with that of $U$ and $V$. However, I do not see why openness is something has to be required as this is basically a consequence of snake lemma with several exact sequences.
 A: It's not quite necessary that $U$ and $V$ be open. Rather, they should be submanifolds whose interiors cover $M$. (And if we work in a more general cohomology theory, we just need any subspaces whose interiors cover.) The reason for this is that we want the short exact sequence of chain complexes $C^*(U\cap V)\leftarrow C^*(U)\oplus C^*(V)\leftarrow C^*(U+V)$ (which exists regardless of $U$ and $V$) to yield a long exact sequence that's actually interesting to us, namely $H^*(U\cap V)\leftarrow H^*(U)\oplus H^*(V)\leftarrow H^*(M)$. 
EDIT: By $C^*(U+V)$ I mean cochains which can be decomposed into a sum of a cochain supported on $U$ and one supported on $V$. When $U$ and $V$'s interiors cover $M$, it's immediate from existence of partitions of unity that this is just $C^*(M)$, but otherwise we have no such guarantee.
But if the interiors of $U$ and $V$ don't cover $M$, there's no reason to have $C^*(U+V)$ chain homotopy equivalent to $C^*(M)$, so that we just get a lame shadow $H^*(U\cap V)\leftarrow H^*(U)\oplus H^*(V)\leftarrow H^*(U+V)$ of the long exact sequence we really want. 
All of this remains a bit obscure in de Rham cohomology, because there aren't such wild coverings of manifolds by submanifolds: you can't cover a manifold with finitely many submanifolds of positive codimension. But in general this issue is quite significant. For one example, we could cover the circle with its irrational points $U$ and rational points $V$ (under a bijection with the interval) and from this deduce that $S^1$ has vanishing first cohomology!
