I found this proof online HERE page 43:
Let $z\in V$, then we can write it as:
$z=x+y$ where $x\in Im(P_i)$ and $y\in Ker(P_i)$, where $P_i$ is an orthogonal projection. Then, we have:
$P_i(z)=P_i(x)+P_i(y)=P_i(x)+0=x$ (since x was already in the image and for projectrion we know that $P_i^2=P_i$.)
Therefore,
$P_j(P_i(z))=P_j(x)$ where $i\neq j$. And here, it says that $P_j(x)=0$ with no explanation (and hence proves the claim.). This is what I don't understand. I can pictorially see it (for usual orthogonal projections in $\mathbb{R}^3$, but I would appreciate it if someone would give a hint that why $P_j(x)=0$ in general. It seems to be equivalent to $Im(P_i)\subset Ker(P_j)$ I guess.)
The theorem you cite assumes that $P = P_1 + … + P_n$ is orthogonal and all $P_k$ are orthogonal, so $$\langle x,x \rangle \boldsymbol{\geq} \langle P(x), x \rangle = \langle P_1(x) + … + P_n(x), x \rangle $$ $$= \sum_{k=1}^n \langle P_k(x), x \rangle = \sum_{k=1}^n \langle P_k(x), P_k(x) \rangle$$ $$ \boldsymbol{\geq} \langle P_i(x), x \rangle = \langle x, x \rangle$$ which implies $P_j(x)=0$ for all $j \neq i$. The first inequality comes from theorem 2.16 on page 41 and the equality in the second line is because a projection is orthogonal if and only if it is self adjoint.