Could you help me to check my proof:
let $\{x^k\}$ be a sequence in such set, we use Cantor's diagonal argument to show the existence of convergent subsequence.
There exists a subsequence $\{x^{\sigma_1(k)}\}$ ($\sigma_1:\mathbb{N}\rightarrow \mathbb{N}$, strickly increasing) such that the first coordinate $\{x^{\sigma_1(k)}_1\}$ converges in $\mathbb{R}$. We can take a further subsequence $\{x^{\sigma_2\circ \sigma_1(k)}\}$ converges at the second coordinate $\{x^{\sigma_2\circ \sigma_1(k)}_2\}$, etc. Now we take the diagional sequence $$\{x^{\sigma_k \circ\cdots \cdot\sigma_1(k)}\}_{k=1}^\infty,$$ this sequence converges at each coordinate, let us call the sequence of coordinates limit $y$. We have $|y_n|\leq \frac{1}{n^2}$, thus $y\in l^1$.
To show $x^{\sigma_k \circ\cdots \cdot\sigma_1(k)} \rightarrow y$ in $l^1$, we have that $$\| x^{\sigma_k \circ\cdots \cdot\sigma_1(k)} - y\| = \sum_{n=1}^\infty \bigg|x^{\sigma_k \circ\cdots \cdot\sigma_1(k)}_n - y_n\bigg|,$$ break the sum at some $N$ such that $\sum_{n\geq N} \frac{1}{n^2}\leq\epsilon$ $$\| x^{\sigma_k \circ\cdots \cdot\sigma_1(k)} - y\| = \sum_{n<N} \bigg|x^{\sigma_k \circ\cdots \cdot\sigma_1(k)}_n - y_n\bigg| + \sum_{n\geq N}^\infty \bigg|x^{\sigma_k \circ\cdots \cdot\sigma_1(k)}_n - y_n\bigg|.$$ The second part is less than $\sum_{n\geq N} \frac{2}{n^2} \leq 2\epsilon$, the first part goes to zero as $k\rightarrow \infty$.
Or you could just use the fact that this is the image of the product space $[-1,1]^\mathbb N$ (which is compact by Tychonoff) under a continuous map $$ T: (x_1,x_2,x_3, \ldots) \to ( x_1, x_2/2^2, x_3/3^2, \ldots)$$