$d(x,y) = |f(x) - f(y)|$ on $\mathbb{R}$
Yes, you are right: $d$ is a metric if and only if $f$ is injective and $(\mathbb R,d)$ is bounded if and only if $f$ is bounded.
Assuming now that $f$ is injective, then $(\mathbb R,d)$ and $f(\mathbb R)$ (endowed with the usual metric from $\mathbb R$) are isometric. Therefore
- $(\mathbb R,d)$ is complete if and only if $f(\mathbb R)$ is complete;
- $(\mathbb R,d)$ is compact if and only if $f(\mathbb R)$ is compact.