You don't need to show $x_n$ converges, you need to show a subsequence of $x_n$ converges. The entire question could be rephrased as "show $K$ is a compact subset of $\ell_2$".
First note that $K$ is a bounded set, if $x \in K$ then $\lVert x \rVert_2 \leq \sum_k \frac{1}{k^2} < \infty$. Then by applying a diagonalisation trick, given a sequence $(x_n) \subset K$, we can find a subsequence $(x_{n_m})$ which converges in every component, meaning
$$ x_i = \lim_{m \to \infty} x_{n_m, i}$$
exists for all $i$. This is a common trick when working with infinite sequences, if you're unfamiliar with this argument please say.
It should be clear that $x_i \leq \frac{1}{i}$ hence $x = (x_1, x_2, \ldots) \in K$. So we need to prove $\lVert x_{n_m} - x \lVert_2^2 \to 0$ to prove convergence in $\ell_2$. To do this we exploit the definition, $\sum_n \frac{1}{n^2} < \infty$ so for all $\epsilon > 0$ we can pick $I$ such that $\sum_{i \geq I} \left(\frac{2}{n}\right)^2 < \frac{1}{2} \epsilon$. Then
\begin{align*}
\lVert x_{n_m} - x \rVert_2^2
&= \sum_{i=1}^{\infty} \lvert x_{n_m, i} - x_i \rvert^2\\
&\leq \sum_{i=1}^I \lvert x_{n_m, i} - x_i \rvert^2 +
\sum_{i=I}^{\infty} \left(\frac{2}{n}\right)^2 \\
&< \sum_{i=1}^I \lvert x_{n_m, i} - x_i \rvert^2 + \frac{1}{2}\epsilon \\
&\to \frac{1}{2} \epsilon \quad \text{as $m \to \infty$}
\end{align*}
We could safely interchange the limit $m \to \infty$ and the sum since the sum is now a finite one. This result holds for all $\epsilon > 0$ hence $\lim_{m \to \infty} \lVert x_{n_m} - x \rVert_2^2 \to 0$ as required.
