 Solution

A neighbourhood of $x$ contains an open set containing $x$; $x_n\to x$ when every neighbourhood eventually contains every $x_n$. If two metrics have the same topology they plainly have the same convergent sequences. Conversely, if a $d$-open set $U$ were not $d\prime$-open, some $x\in U$ would have points $x_n\notin U$ with $d\prime (x_n,x)<1/n$; the common-sequence assumption would imply $d(x_n,x)\to 0$, contradicting $d$-openness. Symmetry finishes the proof. This fails generally: on an uncountable set the discrete and cocountable topologies differ, but in both every convergent sequence is eventually constant.