Suppose I'm working in $\mathbb{P}^n$ and I have an irreducible algebraic variety $X$ of dimension $n-d-1$. In the Grassmannian of dimension $d$, can I always find an open set $U$ such that none of the subspaces in $U$ intersect $X$?
This seems to be exactly what this comment is saying. But I can't prove it.
It seems like I can fiddle around with projections and show that there are open sets that contain or do not contain specific subspaces (by representing a subspace $V$ by a projection $P_V$ onto that subspace, and considering the vanishing locus of $P_VPP_V=P$ or $PP_VP=P_V$, where $P$ is the projection onto the subspace in the Grassmannian). But this doesn't seem to help unless I can constrain $X$ to a specific subspace, which seems to be generally impossible.
The solution I like best proceeds via the incidence correspondence. Briefly, let $$\Sigma = \{ (L,p)\subset G(k,n)\times \Bbb P^n \mid p\in L\}$$ be the subvariety of $G(k,n)\times\Bbb P^n$ which tells you when a point $p$ is in a $k$-plane $L$ (this is called the incidence correspondence). This is a closed subset of $G(k,n)\times \Bbb P^n$ (see here for example), and so it's intersection with $G(n,k)\times X$ is closed. Since the projection $\Bbb P^n_k\to\operatorname{Spec} k$ is universally closed, the map $\pi:G(k,n)\times\Bbb P^n\to G(k,n)$ is closed, so $\pi(\Sigma\cap G(k,n)\times X)$ is a closed subset of $G(k,n)$. Now all you need to do is to show that this is a proper closed subset, which you can do by demonstrating one $k$-plane avoiding $X$.