This problem is Exercise 15 of Chapter 6 of Book of Proof by Richard Hammack.
I proved this exercise in a way that was not described in the textbook.
Proposition. If $b \in \mathbb{Z}$ and $b \nmid k$ for every $k \in \mathbb{N}$, then $b = 0$.
Proof. For the sake of contradiction, suppose $b \in \mathbb{Z}$, $b \nmid k$ for every $k \in \mathbb{N}$ and $b \not = 0$. By the definition of divisibility, $k \not = bn$ where $n \in \mathbb{Z}.$ Hence, $\frac{k}{n} \not = b$. Because $\mathbb{N} \subseteq \mathbb{Z}$, it follows that $ b \not \in \mathbb{Q}$. As $\mathbb{Z} \subseteq \mathbb{Q}$, $b \not \in \mathbb{Z}$. However, we previously stated that $b \in \mathbb{Z}$. It is a contradiction that $b \in \mathbb{Z}$ and $b \not \in \mathbb{Z}$.
I am aware of an easier method, but I am simply wondering if the method I have described is valid.