Corners of the Unilateral Shift

Question

Let $H=\ell^2(\mathbb N)$ and $S\in B(H)$ the unilateral shift associated to the canonical basis.

I want to prove that there is no orthonormal basis of $H$ in which $S$ takes the form $$ \begin{bmatrix} \begin{matrix}1&0\\0&1\end{matrix}&X \\ Y&Z \end{bmatrix}. $$ Alternatively, what I want is to show that there is no rank-two projection $P$ such that $PSP=P.$

I think I have a proof of this, but it uses some obscure facts from dilation theory, and I would be really happy with a quick, dirty argument.

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While writing the question I came up with an argument, so I will be posting it as an answer.

Answer 1

This works for any projection $P$.

Suppose that $PSP=P$. Then $$ PS^*PSP=P=PS^*SP. $$ It follows that $$ 0=PS^*(I-P)SP=[(I-P)SP]^*(I-P)SP,$$ and so $(I-P)SP=0$, that is $$PSP=SP.$$ So $SP=P$; given any $x\in PH$, we have $$ Sx=SPx=Px=x, $$a contradiction since $S$ has no eigenvalues.