Since Grassmannian $Gr(n,m)=SO(n+m)/SO(n)\times SO(m)$ is a homogeneous manifold, you can take any Riemannian metric, and average with $SO(n+m)$-action. Then you show that an $SO(n+m)$-invariant metric is unique up to a constant. This is easy, because the tangent space $T_VGr(n,m)$ (tangent space to a plane $V\subset W$) is $Hom(V,V^\bot)$, and your metric must be $SO(V)\times SO(V^\bot)$-invariant. Such a metric is unique, which follows, e.g. from Schur's lemma.

Since Grassmannian $Gr(n,m)=SO(n+m)/SO(n)\times SO(m)$ is a homogeneous manifold, you can take any Riemannian metric, and average with $SO(n+m)$-action. Then you show that an $SO(n+m)$-invariant metric is unique up to a constant. This is easy, because the tangent space $T_VGr(n,m)$ (tangent space to a plane $V\subset W$) is $Hom(V,V^\bot)$, and your metric must be $SO(V)\times SO(V^\bot)$-invariant. Such a metric is unique (up to a constant multiplier), which follows, e.g., from Schur's lemma.