The Grassmannian of orthogonal projections is compact and connected

Idea. Closed and bounded gives compactness. Being the orbit of a point under the connected group $SO(n)$ gives connectedness.

Step 1: compact. A symmetric idempotent has eigenvalues $0$ or $1$, so $\operatorname{rank}P=\operatorname{tr}P$. Hence $\mathrm{Gr}(k,n)=\{P:\ P^T=P,\ P^2=P,\ \operatorname{tr}P=k\}$ is defined by polynomial equations, so it is closed. It is bounded: $\|P\|_F^2=\operatorname{tr}(P^TP)=\operatorname{tr}P=k$. By Heine-Borel it is compact.

Step 2: it is one orbit. Let $P\in\mathrm{Gr}(k,n)$. By the spectral theorem, $P=UE_kU^T$ for some $U\in O(n)$ (an orthonormal basis of eigenvectors, the first $k$ spanning the image). If $\det U=-1$ and $k<n$, replace $U$ by $U'=UD$ with $D=\operatorname{diag}(1,\dots,1,-1)$: $D$ commutes with $E_k$ (the last diagonal entry of $E_k$ is $0$), so $U'E_kU'^T=UDE_kDU^T=UE_kU^T=P$, and $\det U'=1$. So every $P$ is of the form $UE_kU^T$ with $U\in SO(n)$. (If $k=n$, $P=I$ and the set is a single point.)

Step 3: connected. The map $SO(n)\to\mathrm{Gr}(k,n)$, $U\mapsto UE_kU^T$, is continuous and onto, and $SO(n)$ is connected. Hence $\mathrm{Gr}(k,n)$ is connected.

Remark. $\mathrm{Gr}(k,n)$ is the set of $k$-dimensional subspaces of $\mathbb{R}^n$ (a projection is determined by its image). It is a manifold of dimension $k(n-k)$, and $\mathrm{Gr}(1,n)=\mathbb{RP}^{n-1}$ (previous problem). Note: using $O(n)$ instead of $SO(n)$ in Step 3 would not suffice since $O(n)$ is disconnected.

Lesson. An orbit of a connected group is connected; always check whether you need $SO(n)$ or $O(n)$.