One should be more precise. The statement does not follow merely from the fact that there are infinitely many integers. We shall use some properties of the order relation on the set of real numbers. These properties are immediate from the construction of $\mathbb{R}$. Let $x\in \mathbb{R}$

Existence:

There exists a natural number $n\in \mathbb{N}$ such that $-n\leqslant x < n$ ($\mathbb{R}$ has the Archimedean property). Observe that $\{u\in \mathbb{R}|-n\leqslant u <n\}=\bigcup_{i=1}^{2n} \{v\in\mathbb{R}|i-n-1\leqslant v <i-n\}$. Hence, $x\in \{v\in\mathbb{R}|i-n-1\leqslant v <i-n\}$ for some $i\in \{1,2,\ldots,2n\}$. For such $i$, $\lfloor x \rfloor=i-n-1$

Uniqueness:

Suppose that $k,r \in \mathbb{Z}$ such that $k\leqslant x <k+1$ and $r\leqslant x <r+1$. If $k<r$, then $k+1 \leqslant r$ and we have that $k\leqslant x<k+1\leqslant r \leqslant x<r+1$. This is a contradiction (Because we get x<x). Since exactly one of the statements $k<r$, $k=r$ and $k>r$ is true, WLOG, we conclude that $k=r$.

In essence, the given definition is fine but it can be "simplified" this way: $(k\in\mathbb{Z},\;\; (k\leqslant x<k+1)) \Leftrightarrow (\lfloor x \rfloor = k)$ for all $x\in\mathbb{R}$.