How do we prove that if a nonempty subset $E$ of $\mathbb Z$ is bounded below then it has a minimal element without explicitly making use of the greatest-lower bound property (or lub property)?
A different proof used the glb property of $\mathbb R$ and assumed it had already shown $\mathbb Z \subset \mathbb R$ to say $E$ must have a greatest lower bound.
I was wondering if the proof below is correct if we do not explicitly use(reference) glb property.
Proof.
Let $s$ be a lower bound of $E$ and let $F$ be the set of all lower bounds of E. If $r \in E$ then there must exist an $s \in \mathbb Z$ such that $$s\le r \lt s+1$$ where $s+1 \notin F$.
Since there are no integers between $s$ and $(s+1)$ from the above inequality we can deduce $s=r.$
Therefore, we get $F \ \cap \ E = \{r\}.$ Since $r$ is both an element of $E$ and a lower bound of $E$ it is the minimal element of $E$ as well.
This doesn't make any sense. You give two definitions for $s$, first as a lower bound of $E$ and then as the floor of $r$, but they don't coincide. $6$ is a lower bound of $\{10,11\}$ but if we pick $r=11$ then the value that satisfies the inequality is $s=11$. You have no way of knowing that the floor of $r$ is a lower bound of $E$, and as my example shows it sometimes isn't true.