This came up as a remark:
Remark. If X is a normed vector space, and $T : X \to \ell^1$ a linear homeomorphism, then one can show that there exists a countable sequence $\{x_n\}$ in X such that every $x \in X$ can be uniquely represented as
$$x = \sum_{n=1}^{\infty} a_nx_n$$
with $\sum_{n=1}^{\infty} |a_n| < \infty$.
What I have: The only way I could think of incorporating the operator $T$ was to work backwards as follows:
That is, let $\{x_n\}$ be a sequence in $X$ such that
$$Tx_n = e_n = (\underbrace{0, \dots, 0}_{n-1}, 1, 0, \dots),$$
which is a Schauder basis for $\ell^1$. Then for any $y \in \ell^1$, we can uniquely write it as
$$y = \sum_{n=1}^{\infty} b_nTx_n = \sum_{n=1}^{\infty}b_ne_n, \quad\quad\quad\text{and}\quad\sum_{n=1}^{\infty}|b_n| < \infty.$$
Since $T$ is a linear homeomorphism, it is bijective, and both $T$ and $T^{-1}$ are continuous. Hence, $x = T^{-1}y \in X$, and
$$x = T^{-1}y = \sum_{n=1}^{\infty} b_nx_n.$$
Thus, $\{x_n\}$ forms a Schauder basis in $X$, and so we may conclude that $a_n = b_n$, which gives the result.
Is this how one should go about showing this?