Question: I want to prove $\sup(A)=1$ where $A=\frac{m}{m+n}$ where $m,n$ are Natural numbers.  

1- Every element of $A$ is less than $1$ ( $1$ is min ) so it is bounded above.  

2- let $\epsilon>0$ and consider $ 1-\epsilon$.  

Let $a=\frac{M}{M+N}$ I do now know how to use archemidean pronciple and prove $1-\epsilon$ is not an upper bound.  

I have hints and I have done few steps. I need a clean full proof to learn the whole solution please.

Answer: 
So you want to prove that for any $\epsilon > 0$ that there are natural numbers $m,n$ so that

$1-\epsilon < \frac m{m+n} < 1$

Well.... just tool about....

$1-\epsilon < \frac {m}{m+n}=\frac {m+n}{m+n} -\frac n{m+n}=1-\frac n{m+n} < 1 \iff$

$-\epsilon < -\frac n{m+n} < 0 \iff$

$0 < \frac n{m+n} < \epsilon$. 

......

Need more?

......

Well 

$\iff \frac 1{\epsilon} > \frac {m+n}n= \frac mn + 1$.

Need more?

We can find so $m,n$ if we let $n = 1$ and $m > \frac 1{\epsilon} -1$.

=========

Different tooling around:

For any $\epsilon > 0$ there is an $k\in \mathbb N$ so that $\frac 1k < \epsilon$. (Just let $k > \frac 1 {\epsilon}$).

So we want:

$1 -\epsilon < 1-\frac 1k=\frac{k-1}{k} < \frac m{m+n} < 1$.

Well, that requires that $(m+n)(k-1) < km$.

or $mk +nk -m -n <km$ or

$nk < m+n$.

Let $n$ be anything and $m\ge n(k-1)$.
Your guess is right $sup(A) = 1 $.
We need to prove that:

(1) $\forall m,n \in \mathbb{N} \quad \frac{m}{m+n} \leq 1$ (which is obviously true)

(2)$\forall \varepsilon >0 \quad \exists m,n \in \mathbb{N} $ such that $ \frac{m}{m+n}>1-\varepsilon$

So let $0<\varepsilon < 1$ (otherwise you can choose any $m,n \in \mathbb{N})$.
Set $n = 1$.
$$\frac{m}{m+1}>1-\varepsilon \iff$$
$$ m >\frac{1}{\varepsilon}-1$$ 

Now you use Archimedes prinicple and you get $m^{\*}$ such that $m^{\*} >\frac{1}{\varepsilon}-1 $.Finally you found member from set A to satisfy (2) $m=m^{\*} ,n =1$.