The Grassmannian is smooth, irreducible, and has dimension r(n-r)

By [F1], each point has an affine-space neighbourhood of dimension $r(n-r)$. This proves smoothness in the stated local-chart sense and the dimension assertion. If $r=0$ or $r=n$, there is just one subspace, so all the assertions hold for a point. Assume $0<r<n$ henceforth.

Fix $I_0=\{1,\dots ,r\}$. In its chart write a plane as the row space of $(I_r\mid A)$. For any $r$-subset $J$, retain the identity columns indexed by $J\cap I_0$ and assign the columns indexed by $J\setminus I_0$ bijectively to the remaining standard basis vectors of $k^r$. Complete the other columns of $A$ arbitrarily. The minor indexed by $J$ is then $1$ or $-1$. Thus $U_J\cap U_{I_0}$ is a nonempty open subset of both irreducible affine charts.

Each intersection in step 2.1 is dense in $U_J$, since $U_J$ is irreducible. Hence the closure of the irreducible set $U_{I_0}$ contains every $U_J$, and so is the whole Grassmannian. A closure of an irreducible set is irreducible. The chart isomorphisms in [F1] are for the Plucker-image topology itself (their maps are polynomial minors with regular inverses), so this proves irreducibility in that topology. Together with [F2], it makes the image a projective subvariety.