Absolutely convergent series over all functionals

Let X be a normed vector space. Let (x_n)_{n\ge 1} be a sequence in X, such that for any f\in X^*
\sum_{n=1}^{\infty}|f(x_n)|<\infty.
Then there exists a constant C such that
\sum_{n=1}^{\infty}|f(x_n)|\le C\|f\|, \ \forall f\in X^*.

\textbf{Proof}. Consider the operator T:X^*\to l_1, defined by
T(f)=(f(x_1),f(x_2),\ldots).
Clearly T is well-defined linear operator. Moreover T is bounded. We shall check that using the Closed-graph theorem (X^* is Banach, even if X is not). For this purpose take a sequence (f_k)_{k\ge 1}\subset X^* converging to some f, and let Tf_n\rightarrow p for some p\in l_1. Our aim is to show that Tf=p.
Since Tf_k\rightarrow p (in l_1), we have that
\sum_{n=1}^{\infty}|f_k(x_n)-p_n|\rightarrow 0
as k goes to \infty.
This means that for any n\in\mathbb{N} we have
\lim_{k\to\infty}f_k(x_n)=p_n.
On the other hand, as f_k\rightarrow f, we have that
\lim_{k\to\infty}f_k(x_n)=f(x_n).
This means that f(x_n)=p_n, which implies that Tf=p, which is what we needed.
Now, the boundedness of T implies that there is a constant C such that
\|Tf\|_{l_1}\le C\|f\|
which was to be proved.

\textbf{Comment}. We are tempted to say that the series
\sum_{n=1}^{\infty}x_n
are absolutely convergent, and afterwards, that there is some x\in X which equals this sum (in which case the problem's constraint will be fulfilled). This however, needs not to be true (and the obstacle is not only that X is not required to be Banach space).
Here is a (counter)example. Let X=c_0 and x_n=e_n - the basis unit vectors. Then (c_0)^*=l_1 and for any f\in l_1 we have
\sum_{n=1}^{\infty}|f(e_n)|=\|f\|_{l_1}
so that the problem's constraint is satisfied. However, the series
\sum_{n=1}^{\infty}e_n
are not convergent.
Nevertheless, the problem's conclusion is correct, and the constant is 1.

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