Proving carefully that if $S$ is bounded above then it contains its supremum 
Attempt:
Since $S$ is bounded above, then by completeness, we can find $\alpha = \sup S$. We know that for all $\epsilon > 0$ can find an aelement $x_0 \in S$ so that $\alpha - \epsilon < x_0 \leq \alpha $ . Let $\epsilon = 1 $ and so we have
$$ \alpha - 1 < x_0 \leq \alpha $$
We must show that $\alpha \in S$. Note that $x_0 \leq \alpha < x_0 + 1$
Notice that $\alpha $ must be an integer since it leaves in $\mathbb{Z}$ but $x_0 \leq \alpha < x_0 + 1$ is an integer only if $x_0 = \alpha \in S $
and this completes the proof.
Is this a correct and rigorous solution?
 A: The part "$\alpha$ must be an integer" is not correct, as it is not always the case that if $S \subset S^*$ and is bounded above, $\sup S \in S^*$. For example, by search-substitute all $\mathbb{Z}$ in the lemma with $\mathbb{Q}$ and all "integers" in your proof with "rational numbers",  we will prove an obviously false lemma.
An idea about proving the lemma is that integers are not dense; therefore by using some manipulation we can create a subset of $S$, say it is $S'$, that is both bounded up and below, and therefore has only finitely many members while ensuring $\sup S' = \sup S$. Then it will be easy to prove that $\sup S' = \max S'$.
A: As others have pointed out, you need to show that $\alpha=\sup S$ is an integer to complete your proof. Consider the values $\epsilon=1/n$ for $n=1,2,\dots$ and generate a non-decreasing sequence $x_n\in S$ such that $$\alpha-\frac{1} {n}<x_n\leq \alpha$$ It follows that $x_1=x_2=\dots$ because all of them lie in $(\alpha-1,\alpha]$. Clearly $x_n\to\alpha$ and hence each of $x_n$ equals $\alpha$ so that $\alpha$ lies in $S$. 
A: The proof looks ok as a skelton.
1) You reuse $x_0$ after taking $\epsilon = 1$. You should be clear that you are asserting the existence of an $x_0 \in S$. Once you change $\epsilon$, what you are calling $x_0$ will also change. 
2) The argument for why $\alpha$ is an integer is not completely clear. Since $x_0 \in S$, we know $x_0 \in \mathbb{Z}$. Why can't $\alpha = \frac{x_0}{2}$, for example? Why is the only possibility that $\alpha = x_0$? Surely if $x_0 = \alpha$, half of your biconditional is true, but the second half is yet to be verified.
