I want to prove the following:
Be $z$ an upper bound of a subset $S$ of a partially ordered set. It holds: $$\forall \epsilon > 0: \exists a \in S: a > z - \epsilon$$ Hence: $z$ is the least upper bound of $S$.
Proof: We assume the opposite. That means it exists an upper bound $y$ of $S$ with $y < z \Longleftrightarrow z-y > 0$ Let $\epsilon := z-y > 0$. Hence: $$\exists a \in S: a > z - \epsilon = z - z + y = y$$ Hence $y$ is no upper bound of $S \quad \square$.
Well, I think I am missing something quite obvious here, since I am not able to prove this without a contradiction. How could I prove $$\forall \epsilon > 0: \exists a \in S: a > z - \epsilon \implies z \text{ is the least upper bound}$$ directly?
To prove it directly, i propose the following :
Be $z'$ an upper bound, and suppose $z' \leq z$. As $z'$ is an upper bound, $\forall x \in S, x \leq z'$. This implies that for any $\epsilon > 0$, $z' > z - \epsilon$, and thus $z' \geq z - \lim_{ \epsilon \rightarrow \epsilon} 0 = z$. Finally, $z = z'$.
This almost the same proof, except you don't make a false supposition, you leave open the possibility of what you want to show, and you show the only possible issue is this. Many proofs by contradiction can be rewriten this way, and I don't think you're missing anything here.