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.
Best Answer
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).