Here is a great solution given by one of my geometer friends:
    
Take $X=S^2$, carved out by the equation $x_1^2+x_2 ^2+x_3^2=1$ in $\mathbb R^3$, and cover it by the three open strips $U_i=\{(x_1,x_2,x_3)\in S^2\vert \vert x_i \vert \leq \frac 35 \}$.    
1) The $U_i$'s do cover $S^2$: a point on the unit sphere can't have its three   
 coordinates  $\geq \frac 35$ .
2) For all $i\neq j$ we see, by projecting on the coordinate planes, that  $U_i\cap U_j$ is the disjoint sum of two antipodal open spherical quadrilaterals homeomorphic to squares, so that $\mathcal H^1(U_i\cap U_j)=0$.
3) Each $U_i$ deformation retracts to its central great circle,  so that  each $H_{DR}^1(U_i)=\mathbb Z$.
   
It is then clear that given arbitrary nonzero de Rham cohomology classes $0\neq [\omega_i]\in H_{DR}^1(U_i)$ the gluing condition is vacuously satisfied because of (1).   
Nevertheless these cohomology classes can't be glued to a cohomology class in $S^2$ since $\mathcal H^1(S^2)=0$.