I have included a photo of the relevant page from my textbook at the bottom of this post for context ("It is an easy exercise..." in the first paragraph), but the title of my post describes my intent here.
My attempt:
We consider nonempty intervals $I$ and $J$ such that $\sup (I) = \sup (J)$ and $\inf (I) = \inf (J)$. We may as well assume that neither interval contains either endpoint, because such an assumption would mean that both intervals contain that particular endpoint, which doesn't complicate the business of showing that the two intervals are equal at all.
Consider $x \in I$. Since $\inf (I) < x < \sup (I)$, Proposition 10.5 on page 95 of my text tells me that there exist $x', x'' \in I$ such that $\inf (I) < x' < x < x'' < \sup (I)$.
We would like to show that $x', x'' \in J$, because that will imply that $x \in J$. Because $\inf (J) < x' < x'' < \sup (J)$, Proposition 10.5 tells us again that there exist $y, y' \in J$ such that $\inf (J) < y < x' < x'' < y' < \sup (J)$, which implies that $x', x'' \in J$, which in turn implies that $x \in J$, and therefore $I \subseteq J$. The proof that $J \subseteq I$ proceeds in an entirely symmetric fashion.
I appreciate any responses.
Context:
(Analysis I, Amann and Escher, p. 100)