= Solution
For every finite scalar sequence $a=(a_n)$ with $|a_n|\leq1$, define a bounded linear functional on the <continuous dual space> $X^*$ by
$$
T_a(f)=\sum_na_nf(x_n).
$$
For each fixed $f\in X^*$, the hypothesis gives
$$
\sup_a|T_a(f)|
=\sum_{n=1}^{\infty}|f(x_n)|<\infty,
$$
where the supremum is over all finite choices and the phases $a_n$ may be chosen to align the complex numbers $f(x_n)$.
The dual $X^*$ is a <Banach space> even when $X$ is incomplete. The <Uniform boundedness principle> therefore gives a constant $C$ such that
$$
\sup_a\lVert T_a\rVert\leq C.
$$
For each $N$, choose $a_n$ for $n\leq N$ so that $a_nf(x_n)=|f(x_n)|$. Then
$$
\sum_{n=1}^N|f(x_n)|
=|T_a(f)|
\leq C\lVert f\rVert.
$$
Letting $N\to\infty$ proves
$$
\boxed{\sum_{n=1}^{\infty}|f(x_n)|\leq C\lVert f\rVert
\quad\text{for every }f\in X^*.}
$$
This is the <uniform bound from pointwise absolute summability of dual evaluations>.
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