In Exercise 9, section 4 on page 35 of Munkres’ Topology 2E, the problem is stated as follows.
Exercise 4.
(a) Show that every nonempty subset of $\mathbb{Z}$ that is bounded above has a largest element.
(b) If $x\notin\mathbb{Z}$, show there is exactly one $n\in\mathbb{Z}$ such that $n<x<n+1$.
(c) If $x-y>1$, show there is at least one $n\in\mathbb{Z}$ such that $y<n<x$.
(d) If $y<x$, show there is a rational number $z$ such that $y<z<x$.
I have attempted the solution and the following is my attempt.
Proof:
(a) Let $A$ be the set of integers $k$ such that every nonempty subset of $S_{k+1}=\{1,\ \ \ldots\ ,k\}$ that is bounded above has a largest element. We shall prove that $A\ =\mathbb{Z}_+$.
Firstly, it is clear that $1\in A$ since the set $S_2=\{1\}$ has only a single nonempty subset, i.e., $\{1\}$ itself, in which 1 is clearly the largest element. Now suppose that $n\in A$ so that every nonempty subset of $S_{n+1}=\{1, \dots\ ,\ n\}$ has a largest element.
Consider any nonempty subset $B$ of $S_{n+2}=\{1,\ \dots\ ,\ n+1\}$, noting that $S_{n+2}=S_{n+1}\cup\{n+1\}$. If $n+1\in\ B$, then, then it is clear that $n+1$ is the largest element of $B$. If $n+1\notin \ B$, then clearly $B\subset S_{n+1}$ so that $B$ has a largest element by the induction hypothesis.
Hence in either case $B$ has a largest element so that $n+1\in A$. This shows that $A$ is an inductive set of positive integers so that $A\ =\mathbb{Z}_+$ as desired by the Principle of Induction.
It is obvious that every nonempty subset of $\mathbb{Z}_-\cup\{0\}$ has a largest element. It follows that every nonempty subset of $\mathbb{Z}=\mathbb{Z}_+\cup\left\{0\right\}\cup\mathbb{Z}_-$ that is bounded above has a largest element.
(b) Suppose $x\in\mathbb{R}$ and $x\notin\mathbb{Z}$, and let $C=\left\{k\in\mathbb{Z} \right| k<x\}$. Clearly, there is an $m\in\mathbb{Z}$ where $m<x$ since $\mathbb{Z}$ has no lower bounds; hence by definition $m\in C$ so that $C\neq\emptyset$. Clearly $C$ is bounded above since $x$ is an upper bound of $C$. It then follows from part (a) that $C$ has a largest element $n$ and $n<x$.
Now, suppose for the moment that $n+1\le x$. Then, since $\mathbb{Z}$ is inductive $n\in\mathbb{Z}$ implies that $n+1\in\mathbb{Z}$ too. But $x\notin\mathbb{Z}$ so that it must be that $n+1\neq x$, and hence $n+1<x$. Then $n+1\in A$ so that $n+1\le n$ since $n$ is the largest element of A, contrary to the obvious fact that $n+1>n$, hence we must have $n+1>x$ and thus we have shown that there is exactly one $n\in\mathbb{Z}$ such that $n<x<n+1$.
(c) Let $x-y>1$. Suppose there is no $n\in\mathbb{Z}$ such that $y<n<x$, i.e., there is no integer between $x$ and $y$. Then, $x-y\le1$, contrary to the fact that $x-y>1$. Thus, there must be at least one $n\in\mathbb{Z}$ such that $y<n<x$ if $x-y>1$.
(d) Let $y<x$. If $x-y>1$, by part (c), there is at least one $n\in\mathbb{Z}$ such that $y<n<x$. Then, $n$ is the required rational number.
If $x-y=1$, then there are two possibilities:
(i) both $x$ and $y$ belong to $\mathbb{Z}$, in this case, there is a rational $z$ such that $y<z<x$, and
(ii) neither $x$ nor $y$ does not belong to $\mathbb{Z}$, in this case by part (b), there are three integers $m_1,m_2,\ m_3$ in $\mathbb{Z}$ such that $m_1<y<m_2<x<m_3$. Then, $m_2$ is the required rational number.
If $x-y<1$, then choose an integer $N\in\mathbb{Z}$ such that $Nx-Ny>1$, then by part (c), there is an integer $M\in\mathbb{Z}$ such that $Nx<M<Ny\Longrightarrow x<\frac{M}{N}<y$. Then $\frac{M}{N}$ is the required rational number.\qed
I wonder if there is any error in my attempt to the solution or not. If there is any, please correct it.