I am continuing my work on Lindstrom's real analysis book Spaces (those who have answered my past questions know this by now), and I am working on exercise 2.3.10 which asks for a proof of a special case of the Heine-Borel Theorem.
The Heine-Borel Theorem: Assume that $\mathcal{I}$ is a family of open intervals such that $[0, 1] \subseteq \bigcup_{I \in \mathcal{I}} I$. Then, there is a finite collection of intervals $I_1, I_2, \dots, I_n \in \mathcal{I}$ such that $[0, 1] \in \bigcup_{k = 1}^n I_k$.
The question first guides us to define the set$$A := \big\{x \in [0, 1]: \text{there exists} \{I_k\}_{k=1}^n \subseteq \mathcal{I} \text{ such that } [0, x] \subseteq \cup_{k = 1}^n I_k\big\}.$$
Then, it asks us to show that $c = \sup A$ is finite and positive (Since $0 \in A$ and $1$ is an upper bound for $A$, $c$ is finite. Also, since $0 \in (\alpha, \beta)$ for some $\alpha, \beta \in [0, 1]$, we must have $\beta \in A$ since $\beta \in [0, 1]$ is contained in some open interval that must overlap with $(\alpha, \beta)$.).
For Part (b), the question asks us to show that $c = 1$, but I am having some trouble with that. If we go by contradiction and assume $c < 1$, then there exists some $x \in (c, 1]$ such that no finite collection of intervals covers $[0, x]$. I am not sure where to take it from there. I thought that the $\varepsilon$ characterization of the supremum might help here (for all $\varepsilon > 0$, there exists $a \in A$ with $c - \varepsilon < a$), but I don't quite see how that helps us. Any guidance would be greatly appreciated.
