Bounded sequences that form compact sets or not 
a) Give an example of a bounded closed subset of 
  $$ A = \{(x_n) \in \ell^1: \sum_{n\geq1} x_n = 1\}$$
  which is not compact. The metric we consider on A is induced by the normal norm on $\ell^1$.
b) Show that the set 
  $$B = \{(x_n)\in \ell^1: \sum_{n\geq1} n|x_n| = 1\}$$
  is compact in $\ell^1$.

My try:
a)The sequences in $\hat{A} = \{e_i = (0,\ldots 0 , 1 ,0 \ldots, ),\; \forall i \in \mathbb{N}\}$ has no converging subsequences. Is it a closed subset? It seems like it contains all its boundary points (A is not vector space).
b) If we take a sequence $(x^k) \in B$ we also have $(x^k) \in \ell^\infty$. But since all elements in $\ell^\infty$ has subsequences that converges, this is true for  $(x^k)$?
Edited: Davids method seems to work, both Im trying to get it to work with the following hint.
Hint: You can use without proof the diagonalization process to conclude that every bounded sequence $(x_n) \in\ell^\infty$ has a subsequence $x^{n_k}$ that converges in each component. Moreover, sequences in $\ell^1$ are obviously bounded in their $\ell^1$ norm.
So what I need to show is that the convergence of a sequence does not end up outside B?
 A: (a) The attempt works. To see that $S:=\{e_i,i\in\Bbb N\}$ is closed for the $\ell^1$ norm, let $x\notin S$. There are two index $i$ and $j$ such that $x_ix_j\neq 0$. Let $r:=\min\{|x_i|,|x_j|\}$. Then the open ball of center $x$ and radius $r$ is contained in the complement of $S$. 
(b) The problem is that we have to check that we have convergence in $\ell^1$ for the subsequence. 
As $\ell^1$ is complete, we can check that $B$ is precompact, i.e. given $\delta>0$, we can cover $B$ by finitely many balls of radius $<\delta$. It's equivalent to show both properties hold: 


*

*$B$ is bounded in the $\ell^1$ norm;

*$\lim_{N\to +\infty}\sup_{x\in B}\sum_{k=N}^{+\infty}|x_k|=0$.


Indeed, if a set $S$ is precompact, with $\delta=1$ we get that it's bounded, and $2.$ is a $2\delta$ argument (I almost behaves as a finite set). 
Conversely, assume that $1.$ and $2.$ hold and fix $\delta$. Use this $\delta$ in the definition of the limit to get an integer $N$ such that $\sup_{x\in B}\sum_{n=N+1}^{+\infty}|x_n|<\delta$. Then use precompactness of $[-M,M]^N$, where $M=\sup_{x\in B}\lVert x\rVert_1$. 
Note that this criterion works for $\ell^p$, $1\leqslant p<\infty$.
In our case, each element of $B$ has a norm $\leqslant 1$, and for all $x\in V$, 
$$\sum_{k=N}^{+\infty}|x_k|\leqslant \frac 1N\sum_{k= N}^{+\infty}k|x_k|\leqslant \frac 1N.$$
