Let $d_1,d_2$ and $d_3$ metrics on the metric space $M$. If $d_3\succ d_2\succ d_1$ and $d_1\sim d_3$, then $d_1\sim d_2\sim d_3$.
Edit: Where $d_1\succ d_2$, mean that $d_1$ is more fine that $d_2$, then the application $f_{1,2}:(M,d_1)\to (M,d_2)$, where $f_{1,2}(x)=x$ for all $x\in M$ is continuos. And, if $M$ is a metric space, then two metric $d_1$ and $d_2$ they are said equivalent, i.e., $d_1\sim d_2$, when $f_{1,2}:(M,d_1)\to (M,d_2)$ is a homeomorphism. And finally, the relation $d_1\sim d_2$ define a equivalence relation.
My approach: If $d_3\succ d_2$, then the aplication $f_{3,2}:(M,d_3)\to (M,d_2)$ is continuos and if $d_2\succ d_1$, then the application $f_{2,1}:(M,d_2)\to (M,d_1)$ is continuos. Furthermore $d_1\sim d_3$, implies that $f_{1,3}:(M,d_1)\to (M,d_3)$ is a homeomorphism, i.e, $f_{3,1}:(M,d_3)\to (M,d_1)$ is continuos. Then, now I should prove that $f_{1,2}$ and $f_{2,3}$ is continuos, any idea??.
On the other hand, if $d_1\sim d_3$ define a equivalence relation, then if $d_1\sim d_2$ and $d_3\sim d_2$, then $d_!\sim d_2\sim d_3$.
Note that: $$a \sim b \text{ is equivalent to }( a \succ b \text{ and } b \succ a).$$ And $$\text{if }a \succ b \text{ and }b \succ c, \text{ then }a \succ c.$$
In light of these, the proof should be obvious.