Brezis' exercise 5.25.6: weak convergence of Cesàro mean

Question

I'm trying to solve an exercise in Brezis' Functional Analysis

Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $K$ be a non-empty closed convex subset of $H$. Let $(u_n) \subset H$ such that for each $v \in K$, the sequence $(|u_n-v|)_n$ is non-increasing.

1. Check that the sequence $(d(u_n, K))_n$ is non-increasing. Here $d(u_n, K) := \inf_{v\in K} |u_n-v|$.
2. Prove that the sequence $(\pi_K u_n)_n$ converges to a limit $\ell \in K$. Here $\pi_K :H \to K$ is the orthogonal projection onto $K$.
3. Assume here that $(u_n)$ has a property (P), i.e., whenever a subsequence $(u_{n_k})_k$ converges in the weak topology $\sigma(H, H^*)$ to some limit $u \in H$, then $u \in K$. Prove that $u_n \to \ell$ in $\sigma(H, H^*)$.
4. Assume here that $\bigcup_{\lambda >0} \lambda (K-K) = H$. Prove that there is $u\in H$ such that $u_n \to u$ in $\sigma(H, H^*)$ and that $\pi_K u = \ell$.
5. Assume here that $\operatorname{int} K \neq \emptyset$. Prove that there is $u\in H$ such that $u_n \to u$.
6. Let $\sigma_n := \frac{1}{n} \sum_{i=1}^n u_i$. Assume that $(\sigma_n)$ has the property (P) (defined in (3.)). Prove that $\sigma_n \to \ell$ in $\sigma(H, H^*)$.

Could you check if my attempt on (5.) and (6.) contains any logical mistakes? Is there an alternative shorter proof to (6.)?

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5. First, we assume $K$ is the closed unit ball. If $u_m \in K$ for some $m \in \mathbb N$, then $u_n = u_m$ for all $n \ge m$. Let's assume $u_n \notin K$ for all $n \in \mathbb N$. Then $\pi_K u_n = u_n / |u_n|$. It's clear that $(|u_n|)_n$ converges. By (2.), $(\pi_K u_n)_n$ converges. Then $(u_n) = (|u_n| \pi_K u_n)_n$ converges.

Now we come back to the general $K$. There is $x\in \operatorname{int} K$. Then there is $r>0$ such that $\overline{B(x, r)} \subset K$. WLOG, we assume $x=0$. WLOG, we assume $r=1$. Then we come back to the above case of closed unit ball. This completes the proof.

6. It suffices to prove that every subsequence of $(\sigma_n)$ has a further subsequence that converges to $\ell$ in $\sigma(H, H^*)$. We fix a subsequence of $(\sigma_n)$ and for ease of notation, we still denote this subsequence by $(\sigma_n)$. Clearly, $(\sigma_n)$ is bounded, so there is a subsequence $(\sigma_{n_k})_k$ of $(\sigma_n)$ and $x \in H$ such that $\sigma_{n_k} \xrightarrow{k \to \infty} x$ in $\sigma(H, H^*)$. Then $x\in K$. Let's prove that $x=\ell$.

Let $y_n := \pi_K x_n$. By (2.), $y_n \to \ell$. Then $\psi_n :=\frac{1}{n} \sum_{i=1}^n y_i \xrightarrow{n \to \infty} \ell$. We have $$ \begin{align} |x-\ell|^2 &= \lim_k \langle x-\ell, \sigma_{n_k} - \psi_{n_k} \rangle \\ &= \lim_k \frac{1}{n_k} \sum_{i=1}^{n_k} \langle x-\ell, u_i - y_i \rangle \\ &= \lim_k \frac{1}{n_k} \sum_{i=1}^{n_k} \langle y_i-\ell, u_i - y_i \rangle + \lim_k \frac{1}{n_k} \sum_{i=1}^{n_k} \langle x-y_i, u_i - y_i \rangle . \end{align} $$

Clearly, $(|u_i-y_i|)_i$ is bounded by some $C>0$. Then $$ \begin{align} & \left | \lim_k \frac{1}{n_k} \sum_{i=1}^{n_k} \langle y_i-\ell, u_i - y_i \rangle \right | \\ = & \lim_k \left | \frac{1}{n_k} \sum_{i=1}^{n_k} \langle y_i-\ell, u_i - y_i \rangle \right | \\ \le & \lim_k \frac{C\sum_{i=1}^{n_k} |y_i-\ell|}{n_k} =0. \end{align} $$

Then $$ |x-\ell|^2 = \lim_k \frac{1}{n_k} \sum_{i=1}^{n_k} \langle x-y_i, u_i - y_i \rangle. $$

Because $y_i = \pi_K u_i$ and $x \in K$, we get $\langle u_i - y_i, x-y_i \rangle \le 0$. It follows that $|x-\ell|^2 \le 0$ and thus $x=\ell$.