Here is a proof: Least upper bound property implies Cauchy completeness. I do understand the construction of the subsequence, but how can I formally conclude that the sequence we constructed is unbounded by definition? The definition should be: $(\forall s\in F^+)(\exists x_{n_k})(|x_{n_k}|>s)$
May be a method is take an extra step to show that $(n\epsilon)$ is unbounded. But once I prove it, how can I use it to conclude that $x_{n_k}$ is unbounded? May I ask for a proof?
It's in the construction of the sequence. We have a subsequence where:
$$x_{n_{k+1}} - x_{n_k} > \epsilon$$
This means that:
$$x_{n_k} > x_{n_1}+(k-1)\epsilon$$
Now suppose we had an upper bound $s$ we only have to know that there's a $k$ such that $x_{n_1} + (k-1)\epsilon > s$ to get a contradiction which is where the Archimedian property comes into play.
We know that there's an $l$ such that $l > (s - x_{n_1})/\epsilon$ ($\epsilon\ne0$). Putting $k = l+1$ we immediately get $x_{n_1} + (k-1)\epsilon > L$ since $\epsilon>0$, which contradicts the assumption that it was bounded so we conclude that the subsequence is unbounded.