This is a question from Apostol Calculus section I.3.12 q1
If $x$ and $y$ are arbitrary real numbers with $x < y$, prove that there exists at least one real satisfying $x < z < y$
I've seen a lot of answers such as this one question here that use the archimedean property of real numbers. tl;dr let $z = \frac{y+x}{2}$
My question is the following: I attempted solving this using a different approach, relying on Theorems I.32 and 34. I'm wondering if the approach is valid, specifically the strategy. (I'm sure there are issues with the details, I'm new to proofs and working on smoothing out the details).
Theorem I.30 "Archimedean Property" If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive integer $n$ such that $nx > y$.
Theorem I.32: Let $h$ be a given positive integer and let $S$ be a set of real numbers.
(a) If $S$ has a supremum, then for some $x$ in $S$ we have$$x > \sup S - h$$
(b) If S has an infimum, then for some $x \text{ in } S$ we have $$x < \inf S + h$$
Theorem I.34: Given two nonempty subsets $S$ and $T$ of $\mathbb R$, such that $$s \leq t$$ for every $s \in S \text{ and every } t \in T$. Then $S$ has a supremum, and $T$ has an infimum, and they satisfy the inequality $$\sup S \leq \inf T$$
My attempt
Let $S \text{ and } T$ be nonempty sets of real numbers such that $x < y$ for all $x \in S$ and all $y \in T$.
Then, arbitrary $y$ in $T$ is an upper bound for $S$, so $S$ has $\sup S$ such that $$\sup S \geq x \text{ for all } x \in S \quad \text{ (1) }$$ Moreover, from the definition of supremum, we have $$x \leq \sup S \lt y \quad \text{ (2) }$$
Same reasoning for $T$. Arbitrary $x$ in $S$ is a lower bound for $T$, so $T$ has $\inf T$ such that $$\inf T \leq y \text{ for all } y \in T \quad \text{ (3) }$$ Again from the definition of infimum we have, $$x \lt \inf T \leq y \quad \text{ (4) }$$
Assume there exists a $z$ such that $x < z$ for all $x \in S$ and $z < y$. for all $y \in T$. Then $z$ is an upper bound of $S$ so $$\sup S < z \quad \text{ (5) }$$ and, by definition of supremum, we have $$x \leq \sup S \lt z \quad \text{ (6) }$$
Additionally, $z$ is a lower bound of $T$ so $$\inf T > z \quad \text{ (7) }$$ and, by definition of infimum, we have $$z \lt \inf T \leq y \quad \text{ (8) }$$
From (8) and (7) we have $$x \leq \sup S \lt z \lt \inf T \leq y$$ which becomes $$x < z < y$$