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 $n_\epsilon$.
$n\ge \alpha$ 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!