Summary. The answer to your question is "No". But it is "Yes" under the additional condition that $\liminf_n n x_n > 0$. All of this follows from work of T. Šalát in the 1960s. You find some details below.
Earlier this morning, I had posted another answer (now deleted). But on my way to the chocolate shop I've realized it answered a different (and much easier) question I must have dreamed of...
In addition, and what is perhaps more interesting, I remembered that Georges Grekos had told me of work by Šalát on the very question in the OP. Here is a reference:
T. Šalát, On subseries, Math. Zeitschr. 85 (1964), 209-225.
There, Šalát proves, among other things, the following:
Theorem. Let $(a_n)_{n \ge 1}$ be a non-increasing sequence of non-negative real numbers such that $a_n \to 0$ as $n \to \infty$ and $\liminf_n n a_n > 0$. If $(\varepsilon_n)_{n \ge 1}$ is a $\{0,1\}$-valued sequence for which $\sum_{n \ge 1} \varepsilon_n a_n < \infty$, then $\lim_n \frac{1}{n} \sum_{k=1}^n \varepsilon_n = 0$.
This is Theorem 1 in Šalát's paper, which also investigates the logical strength of the assumptions made in the previous statement. In particular, Note 2 on p. 211 shows that, at least in general, you can't replace the hypothesis that $\liminf_n n a_n > 0$ in the above theorem with the weaker condition that $\sum_{n \ge 1} a_n = \infty$. For this, Šalát considers the sequence $(a_n)_{n \ge 1}$ defined by letting $a_{n^n + k} := n^{-(n+2)}$ for all $n, k \in \bf N$ such that $0 \le k < (n+1)^n - n^n$.
Incidentally, Šalát mentions that his theorem is actually a generalization of previous results from:
J. Krzyś, A theorem of Olivier and its generalizations, Prace math. 2 (1956), 159-164 (in Polish)
and
L. Moser, On the series $\sum 1/p$, Amer. Math. Monthly 65 (1958), 104-105,
which focus on the case where $a_n = \frac{1}{n}$ for all $n$.
For the record, the "theorem of Olivier" alluded to in the title of Krzyś's paper is the same for which Igor Rivin has provided a reference here (that's why I remembered of this story, I guess).
