Let $T \subset Gr(n,2n)$ be an algebraic family of complex $n$-dimensional vector subspaces in $\mathbb{C}^{2n}$, $n > 1$, and denote the vector space corresponding to a point $t$ of $Gr(n,2n)$ by $v(t)$. Assume that for all $t,s \in T$, $v(t)$ and $v(s)$ intersect non-trivially.
Does it follow that there exists a $2n-1$ dimensional subspace $W$ such that $ v(t) \subset W$ for all $t \in T$?
If yes, how does one find such $W$ for a given $T$? If no, what is the simplest example of such family $T$?
Let me write down why it fails for $n=2$, for larger $n$, a similar argument should work.
Giving curve in $G(2,4)$ is same as giving a 1-dimensional family of lines in $\mathbb{P}^3$. The corresponding 2-dimensional vector spaces intersect non-trivially is equivalent to saying that any two lines in our family intersect. All these 2-dimensional vector spaces are contained in a 3-dimensional subspace is equivalent to saying that the family of lines is contained in a hyperplane. So, take any hyperplane in 3-space, a point not on it and a curve of degree at least 2 in the hyperplane. Then, this curve parametrizes a family of lines, namely the lines joing the point on the curve with our fixed point outside the plane. That is, we take the cone over the curve with the point as the vertex. Then, any two lines meet at the vertex, but the family is not contained in a hyperplane.