Finite-dimensional complex Stiefel and Grassmann manifolds are compact

The map $X\mapsto X^{\ast }X$ is continuous in the real and imaginary coordinates, so its inverse image of the closed singleton $\{I_r\}$ is closed. Also $\mathrm{\parallel }X{\mathrm{\parallel }}_F^2=\mathrm{tr}(X^{\ast }X)=r$ on this set, so it is bounded. The frame $[e_1\cdots e_r]$ shows it is nonempty.

Let $\mathcal{U}$ be any open cover of ${\mathrm{Gr}}_{\mathbb{C}}(r,n)$. By [F3], the sets $q^{-1}(U)$ for $U\in \mathcal{U}$ are open in ${\mathrm{St}}_{\mathbb{C}}(n,r)$ and cover it. Compactness from step 1.1 gives $U_1,\dots ,U_k\in \mathcal{U}$ whose preimages cover ${\mathrm{St}}_{\mathbb{C}}(n,r)$. Surjectivity of $q$ then makes $U_1,\dots ,U_k$ a cover of ${\mathrm{Gr}}_{\mathbb{C}}(r,n)$. Thus the quotient is compact directly from the open-cover definition.

Step 1.1 proves compactness of ${\mathrm{St}}_{\mathbb{C}}(n,r)$ and step 2.1 proves compactness of ${\mathrm{Gr}}_{\mathbb{C}}(r,n)$, as claimed.