I'm working on a homework problem and I'm not sure I'm understanding how to construct the proof. I have done work on it, but it's not coming together. The problem, "Let $A \subseteq (0, 1)$ be a set of positive real numbers with the following property: For any finite subset $F \subseteq A$, the sum of the all of the numbers in $F$ is less than 1. Show that $A$ must be either finite or countable. [Hint: This is a problem about cardinality, but it shows that there is really no meaningful way to consider uncountable series. Show that for each $n \in N$, the set $A_n = A \cap ( \frac{1}{n+1} , \frac{1}{n}]$ must be finite.]"
What I have so far is something like this: Proof By assumption, any $F \subseteq A$ is finite. The definition of $A$ implies that $F \not\subset \mathbb{Z}$ and $F \not\subset \mathbb{N}$ because the open interval $(0,1)$ contains neither. By assumption, $F$ is finite and so $F \subseteq \mathbb{Q}$. (I know see this is faulty reasoning because F is defined to be finite not simply countable.) Since $\mathbb{Q}$ is countable and $F \subseteq \mathbb{Q}$, $F$ is also countable.
However, though I now see I wasn't on the right path, I'm perplexed by the HINT. How can I show that $A_n = A \cap (\frac{1}{n+1}, \frac {1}{n}]$ must be finite? I see that each $A_n$, $\inf A_n$ and $\max A_n$ exist. But since $A\subseteq\mathbb{R}$, how can it be finite?
An error in your statement is that since $F$ is finite, $F$ must be composed of rational numbers. This is not the case; $\frac{\pi}{10},$ for instance, is irrational, and could comprise the entire set $A$ for all we know (all subsets of such a set are either empty or $\frac{\pi}{10}$ itself, and therefore have sum less than one).
To consider that $A_n = A \cap\big(\frac{1}{n+1},\frac{1}{n}\big]$ must be finite, assume that it is not finite. Then take any $n+1$ numbers from $A_n$; call this sequence $\{a_k\}_{k=1}^{n+1}$. Notice its sum is $$\sum_{k=1}^{n+1}a_k > \sum_{k=1}^{n+1} \frac{1}{n+1} =1,$$ contradicting our assumption about finite subsets of $A.$ Finally, consider what set $\cup_{n=1}^{\infty}A_n$ represents.