The following is the exercise #24 in chapter 4 of Stein's Real Analysis.
Let $\{e_k\}$ be an orthonormal set in a Hilbert space $\mathcal H$. If $\{c_k\}$ is a sequence of positive real numbers such that $\sum c_k^2<\infty$, then the set $$A = \left\{ \sum_{k=1}^{\infty} a_ke_k :|a_k|\leq c_k \right\}$$ is compact in $H$.
I refered to this page and below is my trial.
Let $S_0 = \{(a_1,a_2,a_3,...):a_k \in \mathbb C, \sum_{k=1}^{\infty} a_k e_k \in A \}$, $f_n = \sum_{k=1}^{\infty} a_{n,k}e_k \in A$, and $A_1 = \{a_{n,1} \}_n$. $|a_{n,1}| \le c_1 \forall n$, so $A_1$ is a bounded sequence in $\mathbb C$. Then, there is a subsequence $\{a_{n_j,1} \}_j$ such that $\lim_{j\to \infty}a_{n_j,1} = b_1\quad (|b_1| \le |c_1|)$.
Let $S_1 = \{(a_1,a_2,a_3,...):a_1 \in A_1, S_1 \subset S_0 \}$, and $A_2 = \{a_{n,2} \}_n$....
Continuing this process inducively, $S_m$ can be obtained $(m \in \mathbb N)$ so that $S_{m+1} \subset S_m$ and $\lim_{j\to \infty}a_{n_j,k} = b_k \quad (|b_k| \le |c_k|), k = 1,...,m$. $\{a_{n_j,k}\}_j$ is a sequence of kth element of $S_m$.
Let $f_{n_j} = \sum_{k=1}^{\infty} a_{n_j,k}e_k$, and $f = \sum_{k=1}^{\infty} b_k e_k$.
Then, $\Vert f_{n_j} - f \Vert = \Vert \sum_k (a_{n_j,k}-b_k)e_k \Vert \le \sum_k \Vert (a_{n_j,k}-b_k)e_k \Vert = \sum_k |a_{n_j,k}-b_k|$
After then, I wanted to show that $\lim_{j\to\infty} \Vert f_{n_j} - f \Vert = 0$, but I found it difficult to show $$\lim_{j\to\infty}\sum_{k=1}^{\infty}|a_{n_j,k}-b_k| = 0$$
Any comments about my trial, whether my approach is correct or not, some errors if exists, how to finish this, or other better ideas, would be appreciated. Thank you.
The set $A$ can be described in a more convenient way as $$A=\left \{\sum_{k=1}^\infty x_kc_ke_k\,:\,|x_k|\le 1\right \}$$ The rest of the proof is standard but, in my opinion, slightly more transparent due to the modified description of $A.$
Let $u_n:=\sum_{k=1}^\infty x_k^{(n)}c_ke_k$ be a sequence in $A.$ Then $\{x^{(n)}_k\}_{n=1}^\infty$ form a family of sequences indexed by $k,$ all bounded by $1.$ By applying the diagonal method we get a subsquence $n_j$ such that $x^{(n_j)}_k$ is convergent for every $k.$ WLOG we may assume that $x^{(n)}_k$ is convergent for every $k.$ Let $x_k=\lim_n x^{(n)}_k.$ Then $|x_k|\le 1.$ Fix $\varepsilon>0.$ There exists $K$ such that $\sum_{k=K+1}^\infty c_k^2<\varepsilon ^2.$ Hence $$\left \|\sum_{k=1}^\infty x_k^{(n)}c_ke_k-\sum_{k=1}^\infty x_kc_ke_k\right \|\\ \le \sum_{k=1}^K |x_k^{(n)}-x_k|c_k+\left \|\sum_{k=K+1}^\infty x_k^{(n)}c_ke_k\right \|+\left \|\sum_{k=K+1}^\infty x_kc_ke_k\right \|\\ = \sum_{k=1}^K |x_k^{(n)}-x_k|c_k+\left (\sum_{k=K+1}^\infty |x_k^{(n)}|^2c_k^2\right )^{1/2}+\left (\sum_{k=K+1}^\infty |x_k|^2c_k^2\right)^{1/2} \\ \le \sum_{k=1}^K |x_k^{(n)}-x_k|c_k+2\varepsilon$$ Now the first finite sum is less than $\varepsilon$ if $n$ is large enough.