Show that any upper bounded subset of $\mathbb{Z}$ has a unique maximal element
This is what I've managed to do so far.
Let $X \subseteq \mathbb{Z}$ be a nonempty upper bounded nonempty subset of $\mathbb{Z}$. Since $\mathbb{Z} \subseteq \mathbb{R}$ we have that $X$ is upper bounded in $\mathbb{R}$ hence $\sup X$ exists. Let $\alpha = \sup X$. We claim $\alpha = \max X$.
This is where I get stuck, ultimately I have to show that $\alpha \in X$, and I'm not exactly sure how to go about doing that.
Since $\alpha$ is a least upper bound, for any $\epsilon > 0$, there is an $n \in X$ such that $\alpha - \epsilon < n \leq \alpha$. This follows from the definition of a least upper bound.
If $\alpha \not\in X$, we have two cases. If $\alpha$ is an integer, then we immediately get a contradiction using the previous paragraph, since there is no integer between $\alpha - 1/2$ and $\alpha$. If $\alpha$ is not an integer, take $\lfloor \alpha\rfloor$ and consider $\epsilon = \alpha - \lfloor\alpha\rfloor > 0$, and get a similar contradiction, using the first paragraph.