Find a bounded sequence in the sequence space $\ell^2$ such that it has no convergent subsequence.
I think the sequence $\{e_n\}_{n=1}^\infty$ works where $e_1=(1,0,...)$, $e_2=(0,1,0,...)$, ....
I would be glad if you could check my attempt.
Attempt. To get a contradiction, suppose it has a subsequence $\{e_{n_k}\}_{k=1}^\infty$ that converges to some point $x=(x_1,x_2,...)$ in $\ell^2$. Then, we have $$0=\lim_{k\to\infty}\|e_{n_k}-x\|_2=\lim_{k\to\infty}\left(\left(\sum_{i=1}^\infty x_i^2\right)-x_{n_k}^2+(1-x_{n_k})^2 \right)=\lim_{k\to\infty}\left(1-2x_{n_k}+\sum_{i=1}^\infty x_i^2\right)$$ which implies that $$\lim_{k\to\infty}x_{n_k}=\frac{1+\sum_{i=1}^\infty x_i^2}{2}>0$$ So $x\notin\ell^2$, contradiction.
Your solution seems alright. The ${e_n}_k$ notation weirds me out, but that’s a small detail.
Here’s another solution. You could simply say that for all $i$, if $x_i\neq0$, $$\lim_{k\to\infty}||e_k-x||_2\geq\lim_{k\to\infty}\left({e_k}_i-x_i\right)^2=x_i^2>0,$$ so that the limit, if it exists, is the null vector. But in that case, the limit would be $1$, contradiction.