I need to prove that $1$ is the supremum of $S=[0,1)$.
Let $a\in S$. Then by definition of the interval $S$ we have $0\leq a < 1$, in particular, $a<1$, $\forall a \in S$ so that $1$ is an upper bound for $S$. By trichotomy, taken another upper bound $w \in \mathbb{R}$ we either have $w =1$, $w<1$ or $w>1$. Now if $1\leq w$ then we're done. So we want to prove by contradiction that is never the case that $w<1$ and $w$ is an upper bound.
Now suppose that $w$ is the supremum of the set $S$. I.e. $w \geq a$ $\forall a \in S$ so that $w$ is an upper bound, and also $w < 1$ (as $w=1$ is already considered above, so here $w \neq 1$) as $w$ is the least upper bound.
Notice that in particular $w\geq 0$ so joining these two we have $0\leq w <1$, i.e. $w\in S$.
Now since $w<1$, take $\epsilon = \frac{1-w}{2} >0$. Then we have $w<w+\epsilon = w+\frac{1-w}{2} = \frac{1+w}{2}<1$ where $\frac{1+w}{2}\in S$ and so $w$ cannot be the least upper bound.
Hence $w$ is not the supremum and $1$ is the least upper bound.
I am pretty sure this proof that I've provided is wrong. Particularly the part highlighted. I'd like to make a very rigorous proof, however I am stuck there. I would like to use the Archimedean property or any other standard result. I don't wanna use anything from intuition only.
Do you know how I can do that?
EDIT: I don't wanna use the approximation property for supremum! I just want to use the basic stuff
This looks fine. To improve your proof, you could:
Explicitly state that we are assuming, towards a contradiction, that $w \neq 1$ (this allows us to infer that $w < 1$ and not merely that $w \leq 1$).
Explicitly give an $\epsilon$. For example, take $\epsilon = \frac{1 - w}{2}$, which is positive since $w < 1$. Verify for yourself that this choice of $\epsilon$ satisfies $w + \epsilon \in S$.