Here's one way: Note that there is a transitive group action on the Grassmannian. The group $GL(n)$ act transitively on the Grassmannian. So choose a smooth point $P \in \mathbb{G}r(k,n)$. Then by acting on $P$ with $GL(n)$, we conclude that all points are smooth and that the Grassmannian is connected.
Here's one way: Note that there is a transitive group action on the Grassmannian. The group $GL(n)$ act transitively on the Grassmannian. So choose a smooth point $P \in \mathbb{G}r(k,n)$. Then by acting on $P$ with $GL(n)$, we conclude that all points are smooth and that the Grassmannian is connected.
But smooth connected varieties are irreducible.