Here is my observation:
Suppose there is an orthogonal projector $P$ such that $P=P^2$.
Then for arbitrary $x$, $Px$ and $(I-P)x$ are orthogonal.
So we have $$ x^* P^* (I-P)x=0$$ where $A^*$ means a transpose of $A$.
Since $x$ is arbitrary, it implies $$ P^*(I-P)=P^*-P^*P=0$$ Note that substituting $x=I$ yields the same result.
Now suppose that $P$ has SVD $P=USV^*$ where $U$, $V$ are unitary and $S$ is a positive diagonal matrix.
Then $$P^*= VSU^*=VSU^*USV^*=VS^2V^*=P^*P$$ Now by multiplying $S^{-1}V^*$ to the left of both sides, $$U^*=SV^*$$ and by multiplying $U$ to the left of both sides, $$I=USV^*=P$$ which is not obviously true.
I think my proof has a critical error, but I could not find it.
Would you give me an advice?
Thanks.
The problem is the step where you "multiply $S^{-1}V^*$ to the left of both sides". If $S$ is invertible, then that implies that $P = USV^*$ is invertible. However, if a projector $P$ is invertible, then we have $$ P^2 = P \implies P^{-1}P^2 = P^{-1}P \implies P = I, $$ which was the conclusion that you reached. If $P$ is a projector with $P \neq I$, then it must hold that $P$ is singular.