I'm reading my text and the author states "$\ell^1$ is separable" as a theorem and then shows that $\ell^1$ is complete by showing that every absolutely summable series in $\ell^1$ converges, which indeed shows that $\ell^1$ is complete.
However, I must be missing something - does this also show that $\ell^1$ is separable by being complete?
Lemma: A normed space $X$ is complete iff any absolutely convergent series in $X$ converges.
Theorem: $\ell^1$ is complete and separable.
Proof: Suppose $\left\{ x_n \right\} \in \ell^1$ is such that $\sum_n \|\ x_n \|_{\ell^1}\ = s$. Then, we have $$\left| \sum_{i=I}^{J} \sum_{n=M}^{N} a_{ni} \right| \leq \sum_{i=I}^{J} \sum_{n=M}^{N} | a_{ni} | \leq \sum_{n=M}^{N} \|\ x_n \|_{\ell^{1}}\ $$ In fact taking the $i$th column, we can see that the whole array sum is bounded since,
$$\sum_i |a_i| = \sum_i \left| \sum_{n=0}^{\infty} a_{ni} \right| \leq s$$ so that $x := \left\{ a_n \right\}$ belongs to $\ell^1$. Finally, note that any rectangular sum goes to $0$ as it moves downward, because $\sum_{n=N}^{\infty} \|\ x_n \|_{\ell^1}\ \to 0$ as $N \to \infty$. Hence,
$$\|\ x - \sum_{n=1}^{N} x_n \|_{\ell^{1}}\ = \sum_{i=0}^{\infty} \left| a_i - \sum_{n=1}^{N} a_{ni} \right| = \sum_{i=0}^{\infty} \left| \sum_{n=N+1}^{\infty} a_{ni} \right| \to 0$$ giving $x = \sum_{n=0}^{\infty} x_n$.
Take a Schauder basis for any vector $x = \left\{ a_n \right\} \in \ell^1$ so that, \begin{align*} \|\ x - \sum_{n=0}^{N} a_n e_n \|\ &= \|\ (a_0, a_1, \dots ) - (a_0, \dots , a_N, 0, 0, \dots ) \|_{\ell^{1}}\ \\ &= |\| (0, \dots , 0 , a_{N+1} , \dots ) \|_{\ell^{1}}\ \\ &= \sum_{n=N+1}^{\infty} | a_n | \to 0 \hspace{.5cm} \text{as} \hspace{.5cm} N \to \infty \end{align*} since $\sum_{n} |a_n| $ converges. If $x = \sum_{n=0}^{\infty} b_n e_n $ as well, then $b_m = e_m x = a_m$ for each $m \in \mathbb{N}$, so $e_n$ form a Schauder basis and moreover any absolutely convergent series in $\ell^1$ converges, so $\ell^1$ is complete.