Projection of an non-increasing sequence of closed convex subsets of a Hilbert space [Haim Brezis Exercise 5.5] This question comes from the Exercise 5.5 of Haim Brezis' Functional analysis, and a related but unsolved post is here: Prove that for every $f$ in $H$, the sequence $u_n$ which is the projection of $f$ on $K_n$ converges to a limit.
The question can be described as follows:
Let $(K_{n})_{n=1}^{\infty}$ be a family of non-increasing sequence of closed convex set in a Hilbert space $H$ with $K:=\bigcap_{n}K_{n}\neq\varnothing$. The book has ensured that $K$ is also closed and convex. Let $P_{K_{n}}$ be the projection onto $K_{n}$, i.e. the map that to $x\in H$ associate the unique point $y=P_{K_{n}}x\in K_{n}$ such that $$\|x-y\|=dist(x,K_{n})=\inf_{z\in K_{n}}\|x-z\|.$$ This property has been ensured by the Hilbert projection theorem.

Then, show that for all $x\in H$, $$\lim_{n\rightarrow\infty}\|P_{K_{n}}x-P_{k}x\|=0.$$


I have deleted my attempt and some edit because they contain several mistakes. I just wrote a proof, so I will directly post it below.
 A: Please let me know if this proof is wrong, since I am not confident in it.
The idea is to prove that $P_{K_{n}}x$ indeed converges and the limit is $P_{K}x$, so the proof is divided into two steps.
To show convergence, using  the Parallelogram law with $a=x-P_{K_{n}}x$ and $b=x-P_{K_{m}}x$ and $m\geq n$ to obtain $$\Bigg\|x-\dfrac{P_{K_{n}}x+P_{K_{m}}x}{2}\Bigg\|^{2}+\Bigg\|\dfrac{P_{K_{n}}x-P_{K_{m}}x}{2}\Bigg\|^{2}=\dfrac{1}{2}(\|x-P_{K_{n}}x\|^{2}+\|x-P_{K_{m}}x\|^{2}).$$
Then, using convexity of $K_{n}$ and $P_{K_{m}}x\in K_{n}$, you can get $$\|P_{K_{n}}x-P_{K_{m}}x\|^{2}\leq 2\|x-P_{K}x\|^{2}-2\|x-P_{K_{n}}x\|^{2}=2[dist(x, K_{m})]^{2}-2[dist(x, K_{n})]^{2}.$$
Note that the $dist(x, K_{n})$ funciton, if considered as a sequence in $n$, is non-increasing and bounded above. So taking $n,m\rightarrow\infty$, the above estimate shows that $(P_{K_{n}}x)$ is Cauchy, and thus converge, say to $u$.
It is clear that $u\in K$ since the limit of these $K_{n}$ is $K$. (I am not sure how to formalize this).
To show $u=P_{K}x$, note that $$dist(x, K_{n})=\inf_{z\in K_{n}}\|x-z\|\leq \|x-y\|\ \ \text{for all}\ \ y\in K_{n}.$$ Since any $w\in K$ belongs to $K_{n}$, we have  $$\|x-P_{K_{n}}x\|=dist(x, K_{n})\leq \|x-w\|\ \ \text{for all}\ \ w\in K.$$
Passing to the limit, we get $$\|x-u\|\leq \|x-w\|\ \ \text{for all}\ \ w\in K.$$ This means that $\|x-u\|$ is the inf: $$\|x-u\|=\inf_{y\in K}\|x-y\|.$$
By definition this means that $u=P_{K}x$ by uniqueness.
Hence, $P_{K_{n}}x$ converges to $P_{K}x$ as $n\rightarrow\infty$, so $$\lim_{n\rightarrow\infty}\|P_{K_{n}}x-P_{K}x\|=0.$$ (I am not sure if I can directly conclude in such a way).
A: Here is another approach to the problem, due to Sakai [1]. Let $P_n $ be the orthogonal projection onto $K_n$. Let $x_0 \in H$ and let
$$x_n = P_n x_0= P_n P_{n-1} \cdots P_1 x_0, \ \ \ n\ge 1. $$
Notice that $x_{n+1} = P_{n+1} x_{n}$. We wish to show that $x_n \to P_K x_0$.
$\textbf{Lemma}$ (Sakai). Assume there exists $ C>0 $ such that
$$ ||{x_n-x_m}||^2 \leq C \sum_{k=m}^{n-1} ||x_{k+1}-x_k|| ^2, \ \ \ n>m \geq 1. $$
Then $ (x_n) $ is norm convergent.
Proof. Since $ x_{k+1} = P_{k+1} x_k $,  we have that
$$ ||x_{k+1}||^2 + ||x_k - x_{k+1}||^2 = ||x_{k+1}||^2+||x_k||^2 - ||x_{k+1}||^2 =||x_k||^2 .$$
Thus $ (||x_k||) $ is a decreasing and thus converges.  Adding the equalities from $ k=m $ to $ k=n-1 $, we obtain
$$ ||x_m||^2=||x_n||^2 + \sum_{k=m}^{n-1} ||x_{k+1} -x_k||^2 .$$
Hence,
$$  ||x_n-x_m||^2 \leq C (||x_m||^2 - ||x_n||^2)  \xrightarrow{n,m \to \infty}0 .$$
In other words, $(x_n)$ is a Cauchy  sequence. QED
Note that if $P$ is a projection then $ ||x-Px||^2 = ||x||^2-||Px||^2$.
Hence,  for $n>m>1$,
\begin{align*}
 ||x_n-x_m||^2 &= ||x_m||^2 - ||x_n||^2 \\
 & = ||x_n-x_{m+1} +x_{m+1}-x_{m+2} + \cdots -x_n +x_n||^2 - ||x_n||^2 
 \\
 &\leq ||x_m -x_{m+1}||^2 + \cdots + ||x_{n-1} - x_n ||^2 + ||x_n||^2 - ||x_n||^2 
 \\
 &= \sum_{k=m}^{n-1} ||x_{k+1} -x_k||^2.
\end{align*}
So, by the above lemma (for $C=1$), $(x_n)$ converges to some limit $y$. It is easy  to check  that, in fact, $y=P_Kx_0$.
[1] M. Sakai, “Strong convergence of infinite products of orthogonal projec-
tions in Hilbert space,” Applicable Analysis, vol. 59, no. 1-4, pp. 109–120,
1995
