I don't believe you are using the archimedean property in a meaningful way. The idea that there exists a number $x$ with the properties that you describe is trivial, take $x=1$ and $n'=2$. That doesn't help with your proof.
What is more meaningful is that for all positive real numbers $x$ we can find an $n'\in\mathbb N$ (based on $x$) such that $1. Since your proof doesn't make this statement, most of what comes after your use of the archimedean property needs some tweaking.
You do successfully prove that $-1$ is a lower bound. If I were approaching the proof (and it greatly depends on what theorems you have already proven) I would suggest assuming that $y$ is a greater lower bound than $-1$ and deduce that there must be an element of $A$ between $-1$ and $y$, which is a contradiction.
Edit, in response to OP's edit: Your proof actually makes no use of the archimedean property anyway. You invoke it to create this number $-1+x$, but the proof makes no real use of this number anyway. You can do without all of that, just assuming that $y$ is a greater lower bound than $-1$, so $-1 and $\forall x \in A,\, y\le x$.
Now you try to prove that there must be an element of $A$ less than $y$ (which I will leave to you to do). Note that it is not enough to just show that there is some rational number between $-1$ and $y$, the elements of $A$ take a particular form.