Consider the set: $$C = \left\{x = (\xi_1, \xi_2, \xi_3, ...) \in \ell_1: \sum_{k=1}^{\infty} 2^k |\xi_k| = 1\right\}.$$
Is this set sequentially compact (in $\ell_1$ norm)? How can I prove it?
I've already proved that $C$ is closed. I want to prove that all sequence has a Cauchy subsequence, but I don't know how to do it.
For a subset of a metric space to be compact, it is necessary and sufficient that it be complete and totally bounded. That you have shown that it is closed implies completeness by the fact that $l^1$ is complete (closed subset of complete metric space is complete). A subset $A \subset X$ of a metric space is totally bounded if for all $\epsilon>0$ there exists $x_1 \dots x_n \in A$ s.t $\bigcup_{i=1}^n B(x_i, \epsilon)$ covers A. Try giving it a shot from here.