A nice geometric way of endowing a Grassmann manifold with a metric (understood here as a distance, and not directly as a Riemannian metric) is to use the Hausdorff distance for subsets of the round sphere.
Consider $V$ a real vector space of dimension $n$ endowed with an inner product, and let $Gr_k(V)$ be the Grassmannian of $k$-planes on $V$. Let $x,y\in Gr_k(V)$, and denote by $S_x$ and $S_y$ the subspheres of the unit sphere $S_V=\{v\in V:\|v\|=1\}$ defined by intersecting it with the subspaces $x$ and $y$ respectively. Then $S_x,S_y\subset S_V$ are closed subsets and the distance between $x,y\in Gr_k(V)$ can be defined as the Hausdorff distance between these sets: $$dist(x,y)=dist_H(S_x,S_y).$$ Although I have not checked the details,
Note that this probably agreesdistance does not coincide with the symmetric space distance induced byon $Gr_k(V)=SO(n)/SO(k)SO(n-k)$. As explained in the homogeneous Riemannian metric discussedpaper of Neretin in Michor's answer, the other posts and comments; but to mesymmetric space distance is computed in terms of the principal angles between two subspaces (namely, it seems easier to visualizeis the square root of the sum of the squares of these angles); while the above distance defined in terms of the Hausdorff distance of closed sets in the unit sphere measures exactly the largest principal angle between subspaces, ignoring all the other smaller principal angles.