what is the meaning of "sequence which has no convergent subsequence"? 
In a proof of non compactness of the closed unit ball of  $\ell^1$, the author says, the sequence $(e_i)$ with $1$ in the $i$th place and $0$ elsewhere, has no convergent subsequence since $\vert\vert e_i-e_j \vert \vert _1=2$ with $i \neq j$

I don't undetstand what this means. That is, what is the meaning of "the sequence which has no convergent subsequence" ? 
 A: It means that if you create a new sequence out of your initial sequence such that you pick only some terms of the original sequence and leave the others out, the remaining sequence does not converge. In more mathematical terms, if $\{a_n\}$ is a sequence and $\{b_k\}$ is a sequence defined by $b_k=a_{n_k}$ such that $n_1 < n_2 < \cdots < n_k < \cdots$, then $b_k$ is not convergent.
One may also reword the same statement like this:
Suppose that $f: \mathbb{N} \to \mathbb{N}$ is a strictly increasing function and define $b_n = a_{f(n)}$. Then $b_n$ is called a subsequence of $a_n$.
Regarding the proof, the author is trying to show that the closed ball in $\ell^1$ is not sequentially compact. Sequential compactness and compactness defined using open covers are equivalent in metric spaces.
Consider $e_1 = (1,0,0,0,0,0,\cdots)$ and $e_2=(0,1,0,0,0,0,\cdots)$
Then, by definiton of the $\ell^1$ norm, we have:
$$\|e_1-e_2\|_1=\sum_n|(e_1)_n-(e_2)_n|=|1-0|+|0-1|=2$$
If instead of $e_1$ and $e_2$ we had $e_i$ and $e_j$, the idea would still be exactly the same.
