$\!$Hej Erik! Your first question asks why an open Schubert cell has the same closure both in the Zariski and in the classical topology. This is the same as asking why the closure of a cell is a closed subvariety. But each cell is defined by a collection of polynomial equations ($f = 0$) and inequations ($f \neq 0$), and it's easy to check that the closure in the classical topology is obtained by just discarding the inequations. Then it's clear that the closure is a subvariety.

For your second question, this follows because Grassmannians admit an algebraic cell decomposition (i.e. they are the disjoint union of locally closed subvarieties isomorphic to some $\mathbf C^d$), the Schubert cell decomposition. See this question: For which varieties is the natural map from the Chow ring to integral cohomology an isomorphism? and the reference to Fulton's book in Donu Arapura's answer.

$\!$Hej Erik! Your first question asks why an open Schubert cell has the same closure both in the Zariski and in the classical topology. This is the same as asking why the closure of a cell is a closed subvariety. But each cell is defined by a collection of polynomial equations ($f = 0$) and inequations ($f \neq 0$), and it's easy to check that the closure in the classical topology is obtained by just discarding the inequations. Then it's clear that the closure is a subvariety.

For your second question, this follows because Grassmannians admit an algebraic cell decomposition (i.e. they are the disjoint union of locally closed subvarieties isomorphic to some $\mathbf C^d$), the Schubert cell decomposition. See this question: For which varieties is the natural map from the Chow ring to integral cohomology an isomorphism? and the reference to Fulton's book in Donu Arapura's answer.

$\!$Hej Erik! Maybe I'm dense but I don't see what you're getting at with your first question. It's true more generally that if $Z \subset X$ is a closed subvariety of another variety, then the Zariski topology on $Z$ is the same as the restriction of the Zariski topology on $X$ to $Z$. Said differently, closed subvarieties of $Z$ are the same as intersections of closed subvarieties of $X$ with $Z$.

For your second question, this follows because Grassmannians admit an algebraic cell decomposition (i.e. they are the disjoint union of locally closed subvarieties isomorphic to some $\mathbf C^d$), the Bruhat decomposition/Schubert cell decomposition. See this question: For which varieties is the natural map from the Chow ring to integral cohomology an isomorphism? and the reference to Fulton's book in Donu Arapura's answer.

$\!$Hej Erik! Your first question asks why an open Schubert cell has the same closure both in the Zariski and in the classical topology. This is the same as asking why the closure of a cell is a closed subvariety. But each cell is defined by a collection of polynomial equations ($f = 0$) and inequations ($f \neq 0$), and it's easy to check that the closure in the classical topology is obtained by just discarding the inequations. Then it's clear that the closure is a subvariety.

For your second question, this follows because Grassmannians admit an algebraic cell decomposition (i.e. they are the disjoint union of locally closed subvarieties isomorphic to some $\mathbf C^d$), the Schubert cell decomposition. See this question: For which varieties is the natural map from the Chow ring to integral cohomology an isomorphism? and the reference to Fulton's book in Donu Arapura's answer.