Here's my attempt but I don't think it's good enough:
We suppose that $\mathbb{N}$ is bounded from above (we already know it's bounded from below). Let $a_n$ be the sequence of natural numbers, $n\in \mathbb{N}$
Since $\mathbb{N}$ is bounded then $a_n$ is also bounded so there exists a $M>0$ such that
$|a_n|<M$
From the Bolzano-Weierstrass theorem there exists at least one subsequence of $a_n$ that converges to a real number $a$ $(1)$ and since $a_n \in \mathbb{N}$ $\Rightarrow $ $a\in \mathbb{N}$
Let $K$ be the set that contains the limits of the subsequences of $a_n$
$K$ has at least one element from $(1)$ and is also bounded since $a_n$ is bounded
Thus there exists the $\sup\mathbb{N}=s$ and in fact $s\in \mathbb{N}$
Then $s-1$ is not an upper bound of $\mathbb{N}$ and thus there is $n\in\mathbb{N}$:
$s-1<n<s\Rightarrow n+1>s$ though $n+1\in\mathbb{N}$ which means that $s$ isn't an upper bound of $\mathbb{N}$
Contradiction
What do you think any ideas??
Best Answer
If you want to prove that $\mathbb{N}$ is bounded using BW, here is a way that uses it more directly.
Define $a_n = n$. Lets assume $\mathbb{N}$ is bounded. So $a_n$ is bounded. So according to BW, there exists a convergent sub sequence $a_{n_k}$ such that $a_{n_k} \rightarrow L$ where L is a real number. So $a_{n_k}$ is a Cauchy sequence, which gives us a contradiction, since $|a_{n_{k+1}}-a_{n_{k}}|>\frac{1}{2}$. So $\mathbb{N}$ isn't bounded.