TITLE: Can one glue De Rham cohomology classes on a differential manifolds?
QUESTION [15 upvotes]: Let $M$ be a differential manifold and $\mathcal H^k$ the presheaf of real vector spaces associating to the open subset $U\subset M$ the $k$-th de Rham cohomology vector space: $\mathcal H^k(U)=H^k_{DR}(U)$. Is this presheaf a sheaf?
Of course not! Indeed, given any non-zero cohomology class $0\neq[\omega]\in \mathcal H^k(U)$ represented by the closed $k$-form $\omega\in \Omega^k_M(U)$ there exists (by Poincaré's Lemma) a covering  $(U_i)_{i\in I}$ of $U$ by open subsets $U_i\subset U$ such that $[\omega]\vert U_i=[\omega\vert U_i]=0\in \mathcal H^k(U_i)$, and thus the first axiom  for a presheaf to be a sheaf is violated.
But what about the second axiom?
My question:
Suppose we are given a differential manifold M, a covering $(U_\lambda)_{\lambda \in \Lambda}$of $M$ by open subsets $U_\lambda \subset M$, closed differential $k-$forms $\omega_\lambda \in \Omega^k_M(U_\lambda)$ satisfying $[\omega_\lambda]\vert U_\lambda \cap U_\mu=[\omega_\mu]\vert U_\lambda \cap U_\mu\in \mathcal H^k(U_\lambda\cap U\mu)$ for all $\lambda,\mu \in \Lambda$.
Does there then exist a closed differential form $\omega\in \Omega^k(M)$ such that we have for the restrictions in cohomology: $[\omega]\vert U_\lambda=[\omega _\lambda]\in \mathcal H^k(U_\lambda)$ for all $\lambda\in \Lambda$ ?
Remarks

This is an extremely naïve question which, to my embarrassment, I cannot solve.
I have extensively browsed the literature and consulted some of my friends, all brilliant geometers (albeit not differential topologists), but they didn't know the answer offhand.
For what it's worth, I would guess (but not conjecture!) that such glueing is impossible.
If the covering of $X$ has only two opens then we can glue.
This follows immediately from Mayer-Vietoris's long exact sequence
$$\cdots  \to \mathcal H^k(M) \to \mathcal H^k(U_1) \oplus   \mathcal H^k(U_2) \to \mathcal H^k(U_1\cap U_2)\to \cdots$$

Update
My brilliant friends didn't answer offhand but a few hours later, unsurprisingly, they came back to me with splendid counterexamples! See below.

REPLY [3 votes]: Here is the great answer given by another of my brilliant friends:
Let $X$ be $\mathbb C, U_0$ be the open complement in $X$ of the closed disk $\bar D=\{z\in \mathbb C\vert \vert z\vert \leq1 \}$
and add a few open discs $U_1,\cdots, U_n$ of radius $\lt 1$ covering $\bar D$ in order to obtain an open covering  $U_0,U_1,\cdots U_n$  of $X$.
Now let $[0]\neq[\omega _0]\in \mathcal H^1(U_0)\cong \mathbb Z$ be a nonzero cohomology clas and define (no choice here!) $0=[\omega_i]\in \mathcal H^1(U_i)=0$.
The compatibility conditions are trivially satisfied since all intersections $U_i\cap U_j (i\neq j)$ are contractible, so that $\mathcal H^1(U_i\cap U_j )=0$.
Nevertheless we can't glue our cohomology classes $[\omega_i]$ to a global cohomology class $[\omega] \in \mathcal H^1(X)$  since the only global cohomology class on $X$ is $0\in \mathcal H^1(X)=0$, which does not restrict to $[\omega _0]\neq 0\in \mathcal H^1(U_0)$.
Remark
Here too the answer is entirely due to my geometer friend.