Every order interval in $l^1$ is norm compact Let $l^1$ denote the space of sequences $(x_n)\subset \mathbb{R}$ with $\Vert (x_n)\Vert_1:=\sum_{n\geq 1} |x_n|<\infty$. We say that $(x_n^1)\leq (x_n^2)$ whenever $x_n^1\leq x_n^2$ for every $n\in\mathbb{N}$. It is well-known that  $(l^1,\Vert\cdot\Vert_1)$ is a Banach space.
Given $(x_n^1),(x_n^2)\in l^1$ with $(x_n^1)\leq (x_n^2)$ we define the order interval
$$[(x_1^n),(x_2^n)]:=\{ (y_n)\in l^1\colon (x_n^1)\leq (y_n)\leq (x_n^2)\}.$$
I suspect that this set is norm compact. 
Any hint to prove that?  

We may consider the set $S=[0, (x_n)]$ only without loss of generality. 
Let $(y_n^k) \in S$. We want to show that it has a convergent subsequence in $S$.
A short but not elementary proof: The set $S$ is compact in the product topology. Thus, there is a point wise convergent subsequence $(y_n^{k_m})$ with a limit $(y_n)\in S$. By dominated convergence theorem, it converges in $\ell^1$. 
