Hilbert cube is compact

Let $\{u_n\}_{n\in \mathbb N}$ be an orthonormal set in $H$ (Hilbert space).

How prove that the set $\displaystyle Q=\{x\in H :\ x=\sum_{i=1}^{\infty}{c_nu_n}, \ \mbox{where} |c_n|\leq\frac{1}{n} \}$ is compact?

$H$ is a Hilbert space over the complex numbers, $c_n\in\mathbb C$

Thanks for any suggestion.

Idea: Show that $Q$ is sequentially compact using a diagonalization procedure.

Let $(x^k)_{k \in \mathbb{N}} \subseteq Q$ be a sequence of the form

$$x^k = \sum_{n=1}^{\infty} c_n^k u_n$$

with $|c_n^k| \leq \frac{1}{n}$. Since $(c_1^k)_{k \in \mathbb{N}} \subseteq B[0,1]$ and $B[0,1]$ is compact, we can pick a subsquence $k^1(j)$ such that $c_1^{k^1(j)} \to c_1 \in B[0,1]$ as $j \to \infty$. Iterating this procedure, we can find a subsequence $(k^{\ell}(j))_j$ of $(k^{\ell-1}(j))_j$ such that

$$c_n^{k^{\ell}(j)} \stackrel{j \to \infty}{\to} c_n \in B \left[0, \frac{1}{n} \right] \tag{1}$$

for all $n \leq \ell$.

Claim: $(x^{k^j(j)})_{j \in \mathbb{N}}$ converges to $x := \sum_{n=1}^{\infty} c_n u_n.$

Proof: For $\varepsilon>0$ (fixed), choose $N \in \mathbb{N}$ such that $$\sum_{n >N} \frac{1}{n^2} \leq \varepsilon.$$ Then $$\begin{align*} \|x^{k^j(j)}-x\|^2 &= \sum_{n=1}^{N} |c_n^{k^{j}(j)}-c_n|^2 + \sum_{n=N+1}^{\infty} |c_n^{k^{j}(j)}-c_n|^2 \\ &\leq \sum_{n=1}^{N} |c_n^{k^{j}(j)}-c_n|^2 + 4 \varepsilon. \end{align*}$$

Recall that $c_n^{k^j(j)}$ is a subsequence of $c_n^{k^N(j)}$ for $j \geq N$. As, by $(1)$,

$$c_n^{k^N(j)} \to c_n$$

for any $n \leq N$, this means that we can choose $M(n)$ such that

$$|c_n^{k^j(j)}-c_n|^2 \leq \frac{\varepsilon}{N}$$

for all $j \geq M(n)$. Setting $M := \max\{M(1),\ldots,M(N)\}$, we get for $j \geq M$

$$ \|x^{k^j(j)}-x\|^2 \leq 5 \varepsilon.$$

Remark: $B[z,r] \subseteq \mathbb{C}$ denotes the closed ball of radius $r$ centered at $z \in \mathbb{C}$.