Supremum and infimum of $\{\frac{1}{n}-\frac{1}{m}:m, n \in \mathbb{N}\}$ I would like to verify my proof of the following:
Let $A=\{\frac{1}{n}-\frac{1}{m}:m, n \in \mathbb{N}\}$.  I want to show that $-1$ and $1$ are the infimum and supremum respectively.
First I will show that $-1$ is the infimum.  To show this I will first demonstrate that it is indeed a lower bound.  Observe: $$\frac{1}{n}-\frac{1}{m} \geq \frac{1}{n}-1 \geq -1.$$
I now claim that $-1$ is the greatest lower bound.  So, by the archimedean property, there exists an $x \in \mathbb{R}$, $x>0$ and $n' \in \mathbb{N}$ such that $1<n'x$.  Thus, $\frac{1}{n'}<x$ and $$\frac{1}{n'}-1<x-1.$$  From the above it is clear that $\frac{1}{n'}-1 \in A$.  Also, it is clear that $-1<-1+x$. 
Let $y=\inf(A)$.  So we let $$-1<y\leq-1+x.$$  By completeness there exists an $r\in\mathbb{Q}$ such that $$-1<r<y\leq-1+x.$$  Thus $y$ is not the infimum.
I was going to note that the supremum was 1 through the relationship between $$\inf(A)=-\sup(-A).$$
 A: *

*$A=B-B$ where $B=\{\frac{1}{n} :  n \in \mathbb{N}\}$ and $X-Y=\{x-y: x \in X, y \in Y\}$.

*$\inf B=0$, $\sup B=1$. ($\inf B=0$ is the Archimedean property.)

*$\inf (X-Y) = \inf X - \sup Y$ implies $\inf A = \inf B - \sup B = 0 - 1 = -1$.

*$\sup (X-Y) = \sup X - \inf Y$ implies $\sup A = \sup B - \inf B = 1 - 0 = 1$.

A: 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 $1<n' x$. 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 $-1<y$ 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.
