Metric space $(X,d)$ is separable if and only if $(X,m)$ is separable, where $ad(x,y)\le m(x,y) \le bd(x,y)$ for any $x,y \in X$ and $a,b>0$. Given two metric spaces $(X,d)$ and $(X,m)$ be two metric spaces such that there exists $a,b>0$ such that
$$ad(x,y) \le m(x,y) \le bd(x,y)$$
for any $x,y\in X$. Show that $(X,d)$ is separable if and only if $(X,m)$ is separable.
EDIT:
Definition. Let $(X,d)$ be a metric space. A neighborhood of $x$ with a radius $r>0$ is defined by $$B(x,r) = \{y \in X : d(x,y) < r\}.$$
Definition. Let $(X,d)$ be a metric space and $D \subseteq X$. Then $D$ is said to be dense in $(X,d)$ if for any $x \in X$ and for all $r>0$, we have $B(x,r)\cap D \ne \emptyset$.
Definition. A metric space $(X,d)$ is said to be separable if there exists $D \subseteq X$ such that $D$ are countable and dense in $X$.
Attempt:
$(\implies)$
Let $(X,d)$ be separable. Then there exists $D \subseteq X$ such that $D$ are countable and dense in $X$. Since $D$ is dense in $(X,d)$, then for any $s>0$ and for all $x \in X$, we have $B_d(x,s) \cap D \ne \emptyset$. Notice that if $d(x,y)<\frac{r}{b}$ for any $x,y \in X$, then
$$m(x,y) \le bd(x,y) < b\cdot \frac{r}{b} = r.$$
Hence, $m(x,y)<r$ and so, $B_d\left(x,\frac{r}{b}\right) \subseteq B_m(x,r)$.
Thus, $B_d\left(x,\frac{r}{b}\right) \cap D \ne \emptyset$ implies $B_m(x,r) \cap D \ne \emptyset$. Therefore, $D$ is dense in $(X,m)$.
Now, since $D$ is countable in $(X,d)$, then $D$ is countable in $(X,m)$.
Hence, $(X,m)$ is separable, as desired.
For the converse direction, I didn't have an idea yet for find the approach. Any ideas? What about the first direction as above? Thanks in advanced.
 A: Given $(X, d) $ separable .
Suppose, $S\subset X$ countable and $d$-dense.
Claim: $S$ is also $m$-dense.
Choose $x\in X$ ,  then $\forall \epsilon >0 $ $\exists s\in S$ such that $d(x, s) <\epsilon $
Now, $m(x, s)\le b d(x, s) \le b \epsilon$
Hence, $S$ is $m$-dense implies $X$ is $m$-separable.
Converse follows immediately just by using the inequality $ad(x, y) \le m(x, y) $.
Alternative :
$S\subset X \space \quad$ countable and $ d$-dense.
Claim: $S$ is also $m$-dense.
Choose any $x\in X $ , then $\exists (s_n) \subset S $ such that $s_n\to x $ in $(X, d) $.
Then, $m(s_n, x) \le b\space d(s_n, s)\to 0 \text { as $n\to \infty $}$
Hence, $S\subset X$ is countable and dense in $(X, m) $ and thus $(X, m) $ is also separable.
Conversely, suppose $(X, m) $ separable and $S$ is countable and $m$- dense subset of $X$.
Then $x\in X$ , $\exists (s_n) \subset S$ such that $m(s_n, x) \to 0 \text { as $n\to \infty$}$.
Now, $\space d(s_n, s) \le\frac{1}{a} m(s_n, s) \to 0 \text { as $n\to \infty $}$
Hence, $(X, d) $ is also separable.
Alternative:
$\exists b\ge a >0$ such that
$ad(x,y) \le m(x,y) \le bd(x,y)$ implies $id:X\to X $ is bi-lipschitz i.e $id$ is both $d-m $ and $m-d$ Lipchitz.
We only need that $id:X\to X$ is bi-continuous i.e a homeorphism.
Then, $S\subset X$ is $d$ -dense iff $m$-dense.(why?)
In fact a continuous onto map between two metric spaces preserve dense sets.
