Let $X$ be a metric space. Two metrics $d(x,y)$ and $r(x,y)$ are equivalent on $X$ if and only if there exist $m,M >0$ such that $$md(x,y)\le r(x,y)\le Md(x,y), \forall x,y\in X$$
Suppose that the following condition holds:
$$\lim\limits_{k\to \infty} x_k = x\mbox{ in (X,d) }\iff \lim\limits_{k\to \infty} x_k = x\mbox{ in (X,r) } \mbox { (*)}$$
I want to prove or disprove that (*) implies equivalence of the metrics.
I think the statement is true. To prove it, I devised the following approach:
Let $(x_k), (y_k)\in X$ be two sequences that converge to $x$ and $y$, correspondingly, under both metrics, $d$ and $r$. Then both sequences are bounded under both metrics. To this end, let $k\le d(x_n)\le K$ and $l\le d(x_n)\le L$. Then $m:=d(k,l) \le d(x_n,y_n)\le \max\{d(K, l),d(L,k)\}=:M$.
From here on, however, I don't know how to proceed. Would appreciate a hint.
The equivalence of convergence of sequences is related to equivalence in topology (assuming you know about topology!). As it turns out, being equivalent metrics is a strictly stronger condition than the metrics generating the same topology.
As for finding an example, maybe you could think of discrete metric spaces with the discrete metric (i.e. takes $1$ when comparing distinct points)? In such spaces, sequences converge if and only if they are eventually constant. Topologically speaking, all this says is that there is some open neighbourhood around each point that only contains the point (i.e. the point is isolated).
Can you think of an infinite subset of, say, the reals, where this is still true, but in terms of the Euclidean metric, the points are getting closer together? Then you can apply the two metrics (discrete and Euclidean) and obtain equivalence in the sense of (*), but not equivalence in metrics.