Solution: Let $x\in \mathbb{R} $ be given. We need to show that there exists a unique integer $n$ such that $n \leq x \lt n - 1$. Consider the following set \[ A \coloneqq \{ a \in \mathbb{Z} : a \leq x \}. \] Since $\mathbb{Z} $ is not bounded below, so the set $A$ is non-empty. $x$ is an upper bound of $A$, therefore by the least upper bound property of $\mathbb{R} $, the set $A$ has the least upper bound (LUB), say $\alpha $.

Since $\alpha $ is the LUB of $A$, so $\alpha - 1$ can not be an upper bound of $A$, therefore, there exists an integer $n\in A$ such that $n > \alpha - 1$. This implies $n+1 > \alpha $ which implies $n+1 \notin A $. Therefore, \[ n + 1 > x. \] Since $n \in A$ so we must have $n \leq x$. Look at the figure below, we have $\alpha -1 \lt n \leq \alpha \lt n+1.$ We also have $n \leq x \lt x$.

Now it remains to show that the above integer is unique. If there exits $n,m\in \mathbb{Z} $ such that $n \leq x \lt n+1$ and $m \leq x \lt m + 1$. Let $n \leq m$. we have \[ n \lt m \leq x \lt n + 1 \lt m + 1, \] which is a contradiction as there does not exist any integer between $m$ and $m+1$ .