Proving that a function doesn't have a limit given $|a-b| = |f(a)-f(b)|$ for all $a,b\in \mathbb R$

For any $\;a\in\mathbb{R}\;$ it results that

$\begin{align} |f(a)|&\ge|f(a)-f(0)|-|f(0)|=|a-0|-|f(0)|=\\ &=|a|-|f(0)|\;, \end{align}$

hence,

$\lim\limits_{a\to\pm\infty}|f(a)|\ge\lim\limits_{a\to\pm\infty}\big(|a|-|f(0)|\big)=+\infty\;,$

consequently,

$\lim\limits_{a\to\pm\infty}|f(a)|=+\infty\;.$

Hint By the triangle inequality $$|f(x)| \geq |f(x)-f(0)|-|f(0)| = |x-0|-|f(0)|=|x|-|f(0)|$$

Consider $b=0$.

As $a \to \infty,\ |f(a)| + |f(0)| \geq|f(a)-f(0)| = |a-0| = |a| \to \infty$.

Hence as $a \to \infty,\ |f(a)| \geq |a| - |f(0)| \to \infty.$