Proof Verification - Archimedean Property I am self-learning real analysis and learning to write proofs. I am trying to prove the Archimedean Property and would like to check if my attempt at a proof is correct and how to improve my proof writing skills.

Given any number $x\in R$, there exists an $n \in N$ satisfying $n>x$.

My understanding of this statement is that the set of natural numbers $N$ is not bounded above.
(Proof): By contradiction, there exists an $x\in R$,such that $\forall n \in N$,$n \leq x$. $x$ is an upper bound for $N$, so by the Axiom of Completeness $N$ has a least upper bound $\alpha = sup (N)$.
By the approximation property, if $\alpha = sup (N)$ then $\forall \epsilon >0$ ,$\exists n \in N$ such that $\alpha - \epsilon < n \leq \alpha$
$\implies \alpha - \epsilon < n$ 
$\implies \alpha < n +\epsilon$
$\implies \alpha \leq n$
$\implies n \geq \alpha$ which contradicts that $\alpha$ is the least upper bound.
 A: There's a mistake. 
Let $\alpha = \sup(\mathbb{N})$ which exists by the reasons you mentiond. It is true that this means that for all $\varepsilon>0$ there exists $n\in\mathbb{N}$ such that $\alpha-\varepsilon <n\leq \alpha$.
From this you conclude that $\alpha<n+\varepsilon$ which is fine.
However this does not mean that $\alpha\leq n$. 
You can't deduce that because $n$ depends on $\varepsilon$, therefore the usual trick of letting $\varepsilon  = (n-\alpha)/2$ isn't possible (you can't define $\varepsilon$ using a variable $n$ which depends on $\varepsilon$).
Instead you should fix $\varepsilon$. If you choose $\varepsilon=\frac{1}{2}$, then $\alpha<n+\frac{1}{2}$ for some natural number $n\in\mathbb{N}$ which corresponds to $\varepsilon=\frac{1}{2}$. From this you can conclude that $\alpha<n+1$, since $n+1$ is a natural number we get a contradiction to the fact that $\alpha$ is an upper bound.
A: Two mistakes.
For all $\epsilon > 0$ the will indeed exist an $n_\epsilon\in \mathbb N$ so that $\alpha -\epsilon < n_\epsilon \le \alpha$ and $n_\epsilon < \alpha +\epsilon$ but that does not mean $n_\epsilon < \alpha + \epsilon$ for all $\epsilon$.
$n_\epsilon < \alpha + \epsilon$ is only true for that $n_\epsilon$ and that $\epsilon$.  For a different value of $\delta > 0$ it will follow that that is an $n_\delta$ so that $n_\delta < \alpha + \delta$ but $n_\delta$ could be a completely different value than $n_\epsilon$.
Second.
$n\ge \alpha$ does not contradict that $\alpha$ is a least upper bound.  $\alpha$ is a least upper bound, and $n \in \mathbb N$ will mean that $\alpha \ge n$ and we have $n \ge \alpha$.   That's not a contradiction.
......
So here's a hint.
let $0 < \epsilon <1$.  
Let $n_\epsilon$ but the natural number where $\alpha - \epsilon < n_\epsilon \le \alpha$.
Now I'll tell you right off the bat, you will never find a contradiction with $n_\epsilon$.  You can note that $n_\epsilon < \alpha+\epsilon$ if you want but that won't be a contradiction nor will it help you.
You will find nothing wrong with $n_\epsilon$.  
Try to find a different natural number that does cause a contradiction.
Second hint.  Don't bother trying to find a different $\delta > 0$ and a different $n_\delta$ so that $\alpha - \delta < n_\delta \le \alpha$.  If you do that you will find something very important about $n_\epsilon$ vs. $n_\delta$ but it won't be a contradiction.
Third hint:  You have $\alpha -\epsilon < n_\epsilon \le \alpha$.  Try to find an $m\in \mathbb N$ so that $m > \alpha$.  That was your original goal after all.  How does knowing $\alpha - \epsilon < n_\epsilon \le \alpha$ help you find $m$ so that $m > \alpha$?
=====
Fourth Hint:  FORGET ANALYSIS!  How would a five year old answer answer this?
Try it.  Go up to a five year old and ask her "I'm thinking of a real big number.  How do you know that there is a bigger one?"  I bet you she will say the answer that is the utter key to this proof!
