Is this subspace isomorphic to $\ell^1$? Let $X$ be a Banach space. 
Suppose there exists a sequence $(x_n)$ in $X$ such that for all finite $A\subseteq\mathbb{N}$ we have that $\|\sum_{n\in A}x_n\|$ equals the number of elements in $A$. 
Does this imply that the subspace spanned by $\{x_n\}$ is isomorphic to $\ell^1$?

In order that the subspace spanned by $(x_n)$ is isomorphic to $l^1$  it has to be closed. Hence, we have to talk about the closure of the span of $(x_n)$.
If $X$ is reflexive, then this subspace is also reflexive, and cannot be isomorphic to $l^1$.
If $X$ is a Hilbert space (or strictly convex space) then no such sequence can exist.
Also the example by @Jochen ($X=\mathbb R^1$, $x_n=1$) shows that span of $(x_n)$ needs not to be infinite-dimensional.
