Let $(H, \langle \cdot, \cdot\rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $(K_n)$ be an increasing sequence of closed convex subsets of $H$. Fix $f \in H$ and let $u_n := \pi_{K_n} (f)$ be the projection of $f$ onto $K_n$.
Prove that $(u_n)$ converges in $H$.
Could you have a check on my attempt?
Let $K := \overline{\bigcup_n K_n}$. Then $K$ is closed and convex. Let $u := \pi_K (f)$. Let's prove that $u_n \to u$.
Let $d_n := |f-u_n|$. Because $K_{n} \subset K_{n+1}$, we get $d_n \ge d_{n+1}$. Then $(d_n)$ is decreasing. So $(d_n)$ is convergent. Let $m > n$. By parallelogram law, $$ \begin{align} 2(d_n^2 + d_m^2) = 4|f - (u_n+u_m)/2|^2 + |u_n-u_m|^2. \end{align} $$
Notice that $(u_n+u_m)/2 \in K_m$, so $|f - (u_n+u_m)/2| \ge d_m$. Then $2(d_n^2 + d_m^2) \ge 4 d_m^2 + |u_n-u_m|^2$ and thus $2(d_n^2 - d_m^2) \ge |u_n-u_m|^2$. Notice that $(d_n)$ is a Cauchy sequence, so $(u_n)$ is convergent.
We have $|f-u_n| \le |f-v|$ for all $v \in K_n$. Then $|f-\lim_n u_n| \le |f-v|$ for all $v \in \bigcup_n K_n$. Because the map $v \mapsto |f-v|$ is continuous, we get $|f-\lim_n u_n| \le |f-v|$ for all $v \in K$. Clearly, $\lim_n u_n \in K$. This completes the proof.