If $(X,d')$ is totally bounded and $d'$ and $d$ are topologically equivalent then $(X, d)$ is separable I am trying to write something similar to the proof of 

If $(X,d)$ totally bounded then $(X,d)$ separable

but I dont know how to use topological equivalence here. Any help?
 A: So you've got that $(X,d')$ is separable. So there exists a countable dense subset $A\subset X$ (this in fact means (according to the definition of a topology generated by metrics) that each $\varepsilon$-ball with respect to $d'$ contains a point from $A$). Take this subset and show that it's going to be dense if you consider balls with respect to $d$ instead of $d'$. It's easy - if your distances generate the same topology, it means that any ball with respect to $d$ contains some ball with respect to $d'$ which in its turn contains a point from $A$. 
A: If you're allowed to use the topological definition of density instead of the metric one, the proof is quite simple: let $S$ be a dense subset of $(X,d')$ (which must exist, since $(X,d')$ was totally bounded, and hence separable).  Since $(X,d')$ and $(X,d)$ were assumed to be topologically equivalent, they have the same topology, so the identity function $i: (X,d') \to (X,d)$ is a homeomorphism.  It follows that $i(S) = S$ is dense in $(X,d)$, and since we already assumed that $S$ was countable, we are done.
