Problem: Let $A$ be an infinite set of real numbers such that $\vert a_1 + a_2 + \dots + a_k \vert \lt 1$ for every finite subset $\left\{ a_1, a_2, \dots, a_k \right\} $ of $A$. Show that $A$ is countable.
Solution: Consider the set
\[
A_n = A \cap \left( \frac{1}{n}, \infty \right),~ n \in \mathbb{N} .
\]
Since for any finite subset of $A$, the absolute sum is less than $1$, we have $\left\vert A_n \right\vert \lt n$, where $\left\vert A_n \right\vert $ denotes the number of elements in the set $A_n$. Similarly, the set
\[
A_{-n} = A \cap \left( -\infty ,-\frac{1}{n} \right) , n \in \mathbb{N}
\]
has the same property. Note that for any $x\neq 0$, there exists $n\in \mathbb{N} $ such that $x > \frac{1}{n}$ or $x \lt -\frac{1}{n}$. Therefore, any nonzero element of $A$ must be an element of either $A_{n}$ or $A_{-n}$ for some $n \in \mathbb{N} $. Thus,
\[
A \subseteq \{ 0 \} \bigcup_{n \in \mathbb{N} } \left( A_n \cup A_{-n} \right).
\]
The right hand side is a countable union of finite sets and hence countable. Now any subset of a countable set is countable implies that $A$ is countable.