I was trying to prove that the grassmannian is a manifold without picking bases, is that possible?
Here's what I've got, let's start from projective space. Take $V$ a vector space of dimension n, and $P(V)$ its projective space. To imitate the standard open sets when you have a basis, consider a hyperplane $H$. We can form a (candidate open) subset $U_H$ consisting of those lines $L \in P(V)$ such that $L \oplus H = V$.
For the Grassmannian you can proceed similarly, say you want to construct $Gr(d,V)$. Take a subspace H of dimension $c = n - d$, and consider the set $U_H$ of those subspaces $W \in Gr(d,V)$ such that $W \oplus H = V$.
I'm not really sure how to proceed after this. Any hints? the main problem is that $U_H$ should be isomorphic to affine space but I can't seem to cook up the natural candidate for it.