We claim that $f(n) \in \mathbb{Z}$, for any integer $n$. Indeed, write $f(f(f(x))) = f([x]) = [f(x)]$ to derive that $f(n) = [f(n)]$ for any integer $n$, implying $f(n) \in \mathbb{Z}$.

Suppose that for all $a, b \in \mathbb{Z}$ we have $|f(a) - f(b)| < |a - b|$. Then $|f(n + 1) - f(n)| < 1$, for any integer $n$. Since both $f(n)$ and $f(n + 1)$ are integers we derive that $f(n) = f(n + 1)$, hence $f(n) = f(0)$, for any integer $n$. It follows that $n = f(f(n)) = f(f(0)) = 0$, a contradiction.

Notice that $f(0) \ne f(1)$, otherwise $0 = f(f(0)) = f(f(1)) = 1$, false. If $|f(0) - f(1)| \ge |0 - 1|$, we are done; if else, from $|0 - 1| = |f(f(0)) - f(f(1))|$ we obtain $|f(f(0)) - f(f(1))| > |f(0) - f(1)|$. The distinct numbers $f(0)$ and $f(1)$ fulfill the claim.