IMC 2004 · Day 1 · P1

Let $S_n = S \cap \left( \frac{1}{n}, \infty \right)$ for any integer $n > 0$. It follows from the inequality that $|S_n| < n$. Similarly, if we define $S_{-n} = S \cap \left( -\infty, -\frac{1}{n} \right)$, then $|S_{-n}| < n$. Any nonzero $x \in S$ is an element of some $S_n$ or $S_{-n}$, because there exists an $n$ such that $x > \frac{1}{n}$, or $x < -\frac{1}{n}$. Then $S \subset \{0\} \cup \bigcup\limits_{n \in \mathbb{N}} (S_n \cup S_{-n})$, $S$ is a countable union of finite sets, and hence countable.