 Bounded metric transform

The increasing function $\varphi (s)=s/(1+s)$ is subadditive on nonnegative arguments: each summand $b/(1+b)$ and $c/(1+c)$ is at least its numerator divided by $1+b+c$. Thus $\varphi \circ d$ is a metric whenever $d$ is. It is bounded and induces the same topology because $\varphi$ and its inverse near zero preserve convergence to zero.