EDIT: This question is not a duplicate because I am asking how to show that $1$ is the greatest upper bound of $S$ by using the fact that $\mathbb{N}$ is not bounded above.
Let $S=\{\frac{1}{n}-\frac{1}{m}:n,m\in \mathbb{N}\}$ Show that $\sup S = 1$
To do this we must show that $1$ is an upper bound of $S$ and $1$ is the least upper bound of $S$.
To show it is an upper bound we know $$n>0, n>\frac{1}{n}, and \frac{1}{n}\leq 1$$ $$\Rightarrow 0<\frac{1}{n}\leq1$$ $$\Rightarrow -\frac{1}{m}<\frac{1}{n}-\frac{1}{m}\leq 1-\frac{1}{m}$$
Since $1-\frac{1}{m}<1, m\in\mathbb{N}$ we know $\frac{1}{n}-\frac{1}{m} < 1$ so $1$ is an upper bound of $S$.
Now I'm stuck on showing that $1$ is the least upper bound. If we let $v\in\mathbb{R}$ be an upper bound of $S$ we want to show $1\leq v$. I am trying to do this by contradiction.
So we assume $v<1 \Leftrightarrow v-1<0\Leftrightarrow 0 <1-v$.
I have been told I can run into a contradiction by recalling that $\mathbb{N}$ is not bounded above, but I'm not sure how to get to the contradiction.
Suppose there was another upper bound $\alpha$ with $\alpha<1$. Choose a positive integer $N$ large enough that $$\frac{1}{N}<1-\alpha.$$
(The fact that we can choose such an $N$ follows from the fact that $\mathbb{N}$ is not bounded above.)
But now the number
$$\frac{1}{1}-\frac{1}{N}>\alpha,$$
contradicting the fact that $\alpha$ is an upper bound of $S$, since the RHS of that expression is in $S$ ($n=1$ and $m=N$).