Let $\{H_n\}_{n \ge 0}$ be a sequence of Hilbert spaces, with $H_{n+1} \subset H_n$ (Clarification: we assume that $H_{n+1}$ is a subspace of $H_n$) for all $n\ge 0$, and denote
$H_\infty=\cap_{n=1}^\infty H_n$. 

**Claim:** If $P_n$ is the orthogonal projection from $H_0$ to $H_n$, and $P_\infty$ is the orthogonal projection from $H_0$ to $H_\infty$, then for all $x_0 \in H_0$, we have $P_n x\to P_\infty x$ in norm as $n \to \infty$.

Proof: For all $n \ge 1$, write $x_n:=P_n x_0$ and $y_n:=x_{n-1}-x_n$. Then $y_n$ is orthogonal to $H_n$ for all $n \ge 1$, so   the elements of the sequence  $\{y_n\}_{n \ge 1}$ are pairwise orthogonal. Thus Bessel's inequality gives $\sum_{n=1}^\infty \|y_n\|^2 \le \|x_0\|^2$, whence $\{x_n\}$ is a Cauchy sequence in $H_0$,
so it converges to some vector $z$. Since $z \in H_n$ for each $n$, we conclude that $z \in H_\infty$. Also, for all $h \in H_\infty$, we have $$0=\langle x_0-x_n,h\rangle \to \langle x_0-z,h\rangle $$ 
as $n \to \infty$, so $x_0-z$ is orthogonal to  $H_\infty$. Therefore $z=P_\infty x_0$.