I wish to verify a proof; solutions to this exercise are not available.
Lemma: If $S \subset \mathbb{Z}$ is bounded from above, it has a maximum element.
Proof: If $S$ is bounded from above, it has a supremum, say $b$. Suppose $b \notin S$. Since $\mathbb{Z}$ is not dense in $\mathbb{R}$, we can find some $\epsilon>0$ such that there is no integer $k$ satisfying $b-\epsilon<k<b$. Hence if every $n\in S$ satisfies $n<b$, every $n \in S$ also satisfies $n<b-\epsilon$.
This contradicts $b$ being a supremum. Thus the supremum of $S$ must be an element of $S$, namely its maximum.
Theorem: If $x$ is an arbitrary real number, there is exactly one integer $n$ satisfying $n≤x<n+1$.
Proof: Let $P = \{k \in \mathbb{Z} | k≤x\}$. This is bounded from above, say by $x$. Hence it has a maximal element $n$. Moreover, $n+1>x$ for otherwise $n+1 \in P$ which would contradict $n$ being the maximum. The uniqueness of $n$ can be observed by noting that every integer $m$ can be expressed as $m=n+c$ for integer $c$ and also that $n+c≤x<n+c+1$ is untrue for $c≠0$