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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 $\mathcal H^1(S^2)=0$.

Here is the solution obtained by one of my brilliant geometer friends evoked in the question:
Take $X=S^2$, the unit $2$-sphere with equation $x_1^2+x_2 ^2+x_3^2=1$, 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 $i\neq j$ it 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 De Rham $0\neq [\omega_i]\in H_{DR}^1(U_i)$ the glueing condition is vacuously satisfied because of 1.
Nevertheless these cohomolgy classes can't be glued to a cohomology class in $S^2$ since $\mathcal H^1(S^2)=0$.
Important remark
My contribution to this answer is: zero, nada, zilch, que dalle...

Here is the solution obtained by one of my brilliant geometer friends evoked in the question:
Take $X=S^2$, the unit $2$-sphere with equation $x_1^2+x_2 ^2+x_3^2=1$, 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 $i\neq j$ it 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 De Rham $0\neq [\omega_i]\in H_{DR}^1(U_i)$ the glueing condition is vacuously satisfied because of 1.
Nevertheless these cohomolgy classes can't be glued to a cohomology class in $S^2$ since $\mathcal H^1(S^2)=0$.
Important remark
My contribution to this answer is: zero, nada, zilch, que dalle...

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 $\mathcal H^1(S^2)=0$.

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 $\mathcal H^1(S^2)=0$.

Here is the solution obtained by one of my brilliant geometer friends evoked in the question:
Take $X=S^2$, the unit $2$-sphere with equation $x_1^2+x_2 ^2+x_3^2=1$, 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 $i\neq j$ it 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 De Rham $0\neq [\omega_i]\in H_{DR}^1(U_i)$ the glueing condition is vacuously satisfied because of 1.
Nevertheless these cohomolgy classes can't be glued to a cohomology class in $S^2$ since $\mathcal H^1(S^2)=0$.
Important remark
My contribution to this answer is: zero, nada, zilch, que dalle...