Define $B \subset \ell^{2}$ by $$B = \{x \in \ell^{2}: \sum_{n=1}^{\infty}n|x_{n}|^{2} \leq 1 \}. $$ Show that $B$ is compact.
I found this question while studying for an exam. I tried proving that $B$ was sequentially compact by taking an arbitrary sequence $\{ x^{(k)} \}_{k \geq 1} \in \ell^{2}$, and using a diagonalization argument to extract a subsequence $\{ x^{k(j)} \}_{j \geq 1}$ with the property that $$\lim_{j \to \infty}x_{n}^{k(j)} = x_{n} \in \left[-\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}} \right],$$ for each $n \in \mathbb{N}$. I proved that $x = (x_1,\dots,x_{n},\dots) \in B$, but can't find a way to prove that my original sequence converges to $x$ in the $\ell^{2}$ norm. Is this approach correct so far, or would it be easier to show that $B$ is complete and totally bounded?
Here's some guiding line. Suppose we have $(x_{n,j})_{j\ge 1}\in B$ such that for each $n\ge 1$, $$ \lim_{j\to\infty}x_{n,j}=x_n $$ exists. We can show that $(x_n)_{n\ge 1}\in B$, by applying Fatou's lemma, for example. Then we have $$ \sum_{n=1}^\infty n|x_{n,j}-x_n|^2 \le 2\left(\sum_{n=1}^\infty n|x_{n,j}|^2+\sum_{n=1}^\infty n|x_n|^2 \right)\le 4 $$ by Cauchy-Schwarz. Finally, from $$ \sum_{n=1}^\infty |x_{n,j}-x_n|^2\le\sum_{n=1}^{N -1}|x_{n,j}-x_n|^2 +\frac1N\sum_{n=N}^\infty n|x_{n,j}-x_n|^2, $$ we can deduce that $x_{\cdot,j}\xrightarrow{j\to\infty} x$ in $\ell^2$.