Every metric space $(X,d)$ gives rise to a bornology by asserting that $A \subseteq X$ is bounded iff there is an open ball (of finite radius) that includes $A$.
Now just because two metrics induce the same topological space, nonetheless they needn't induce the same bornology. E.g. Let $d$ denote the usual metric on $\mathbb{R},$ and take $d'(x,y) = \min\{d(x,y),1\}.$ Then $d'$ induces the same topology as $d$, but every subset of $\mathbb{R}$ is bounded with respect to $d'$.
Question. Suppose two metrics induce the same bornology. Do they necessarily induce the same topological space?
Under the usual metric on $[0,1]$ every subset is bounded. Under the discrete metric every subset is bounded, too.