Prove that subset of Hilbert space is compact I encountered the following difficulty while reading a proof in Bosq's Linear Processes in Function Spaces. 
Let $(v_j)_{j\geq 1}$ be an orthonormal basis in the real Hilbert space $H$. Let $1=M_1<M_2<M_3<\dotsb$ be a strictly increasing sequence of positive integers, and let $(\ell_k)_{k\geq 1}$ be a positive sequence with $\lim_{k\to\infty}\ell_k=0$. Define the sets
$$
B_k:=\left\{x\in H \;\middle|\; \sum_{j=M_k}^\infty\langle x,v_j\rangle^2\leq\ell_k\right\}
$$
for $k\geq 1$. Prove that the set 
$$
K:=\bigcap_{k\geq 1} B_k
$$
is compact. This type of problem reminds me of the Hilbert Cube. Is this a special case of something more general?
 A: Since identifying $(x_n)_{n \in \Bbb N} \in \ell^2$ with $\sum_n v_n \in H$ is an isometry of Hilbert spaces, assume WLOG that $H = \ell^2$.
A standard approach for these kinds of problems is as follows: for each $j$, let $x^j = (x^{j}_n)_{n \in \Bbb N}$ be an element of $K \subset \ell^2$. Our goal is to extract a convergent subsequence; we do so via a "diagonal" argument.
Extract a subsequence $(x^{j,1})_{j \in \Bbb N}$ of $(x^j)_{j \in \Bbb N}$ for which the sequence $x^{j,1}_1$ of first coordinates converges as $j \to \infty$.  Extract a subsequence $(x^{j,2})_j$ of $(x^{j,1})_j$ for which the sequence $x^{j,2}_2$ of second coordinates converges. Proceed for all $k \in \Bbb N$ to find subsequences $(x^{j,k+1})_j$ of $x^{j,k}$ for which the sequence of $k$th coordinates converges.  Note that for all $k$, the sequence $(x^{j,k})_j \in H$ is such that the first $k$ coordinates all converge. The fact that these subsequences are nested ensures that the coordinate sequence $(x^{j,k}_m)_{j \in \Bbb N}$ will have the same limit for all $k \geq m$.
Consider the sequence $(x^{k,k})_{k \in \Bbb N}$. Verify that all coordinates of this sequence converge, which is to say that $x^{k,k}$ converges pointwise. Using the fact that $(x^{k,k})_k \subset K$, show that $x^{k,k}$ also converges in norm.
