Is this subspace isomorphic to the space $\ell^1$? Let $X$ be a Banach space. 
Suppose there exists a sequence $\{x_n\}$ in $X$ such that 


*

*for all finite $A\subseteq\mathbb{N}$, and

*for all functions $\alpha:A\to\{-1,1\}$,


we have that $\|\sum_{n\in A}\alpha_nx_n\|$ equals the number of elements in $A$. 
Does this imply that the closure of the subspace spanned by $\{x_n\}$ is isomorphic to $\ell^1$?
This question is a modification to this question, where $\alpha\equiv 1$.
 A: Yes this is true.
It is even possible to show more, namely that the closure of the span is isomtrically isomorphic to $\ell^1$.
The main part of the proof will be to show that
$$
 \tag{1}
 \label{eq1}
 \|\sum_{i=1}^n a_i x_i\| = \sum_{i=1}^n | a_i|
$$
is true for all $(a_i)\in \ell^1$ and $n\in\mathbb N$.
We prove \eqref{eq1} by induction in $n$.
We assume that \eqref{eq1} was already shown for $n\in\mathbb N$
and we want to show that it also holds for $n+1$.
Wlog we can assume that $a_{n+1}\geq |a_i|$ for $i=1,\ldots n$
(Otherwise we can reorder the indices so that $|a_{n+1}|$
is coefficient with the largest absolute value,
and $a_{n+1}\geq0$ can be assumed because otherwise we can multiply everything by $-1$).
We define $\alpha_i:=\operatorname{sgn} a_i$.
Then we have
$$
\begin{align}
\|\sum_{i=1}^{n+1} a_i x_i\|
&= 
\|a_{n+1}\sum_{i=1}^{n+1} \alpha_i x_i - \sum_{i=1}^n (\alpha_i a_{n+1}-a_i)x_i\|
\\ &\geq 
\|a_{n+1}\sum_{i=1}^{n+1} \alpha_i x_i\|-\| \sum_{i=1}^n (\alpha_i a_{n+1}-a_i)x_i\|
\\ &=
 |a_{n+1}|(n+1) - \sum_{i=1}^n |\alpha_i a_{n+1}-a_i|
\\ &=
 |a_{n+1}|(n+1) - \sum_{i=1}^n (|a_{n+1}|-|a_i|)
 = \sum_{i=1}^{n+1} |a_i|.
\end{align}
$$
The other inequality follows from the triangle inequality.
Thus \eqref{eq1} holds true for all $n\in\mathbb N, (a_i)\in\ell^1$.
By continuity it can also be shown that \eqref{eq1} holds for $n=\infty$.
Let us define the map
$$
 I:\ell^1\to \overline{\operatorname{span}(x_i)},
 e_n\mapsto x_n.
$$
Then it follows from the above that $I$ is an isometry.
Finally, the property that $I$ is an isometry can be used to show that $I$ is surjective and injective.
