Showing two metrics are equivalent I'm trying to solve the following problem.
Let $d$ and $\rho$ be metrics on the same space $M$, and suppose that there exists a continuous function $\phi : [0, \infty) \to [0, \infty)$ such that: 
1) $d(x, y) = \phi(\rho(x, y))$ for all $x, y \in M$.
2) $\phi(t) = 0$ iff $t = 0$.
3) inf$_{t \geq 1} \phi(t) > 0$.
Prove that $d$ and $\rho$ are equivalent metrics, i.e. define the same topology. 
If we let $\tau_d$ be the topology on $M$ generated by $d$, and similarly for $\tau_{\rho}$, then I have already shown by means of Hausdorff's criterion that $\tau_d \subseteq \tau_{\rho}$. 
However, I am having trouble showing the converse inclusion, that $\tau_{\rho} \subseteq \tau_d$. One hint that I have is to use the Bolzano-Weierstrass theorem somewhere, but I am not exactly sure of the best way to proceed.
Any hints or answers are appreciated!
 A: This answer is a pretty big hint.
Let us start with some notation. Write $B^d_\varepsilon(x)$ for the open $d$ ball or radius $\varepsilon$ around a point $x$, and $B^\rho_\varepsilon(x)$ for the open $\rho$ ball. Also define $l=\inf_{t>1}\phi(t)$ which is bigger than $0$ by your point 3.
To show that $\tau_\rho\subseteq\tau_d$ it is enough that for every $x\in M$ and $\varepsilon > 0$ that we can find a $\delta >0$ such that $B^d_\delta(x)\subseteq B^\rho_\varepsilon(x)$. And of course the smaller $\varepsilon$ is the harder it is so let us pick $\varepsilon$ small enough so that $\phi(\varepsilon) < l$. This of course means that $\varepsilon < 1$ by definition of $l$.
Now let us just guess that $\delta = \frac{l}{2}$ will do the trick. If $B^d_\delta(x)\not\subseteq B^\rho_\varepsilon(x)$ then there is some point $y$ such that $d(x,y) < \delta$ but $\rho(x,y) \geq \epsilon$. Now since $\phi(\rho(x,y))=d(x,y)< \delta< l$ it follows that $\rho(x,y)\leq 1$ by deinition of $l$. 
So we have found a value $a_0=\rho(x,y)$ such that $\varepsilon\leq a_0 < 1$ and $\phi(a_0)< \delta=\frac{l}{2}$. 
Now to complete the proof we should also consider smaller values of $\delta$. For example all the values $\frac{l}{2^n}$. And if for none of these values $B^d_\delta(x)\subseteq B^\rho_\varepsilon(x)$ then you can derive a contradiction with your assumptions about $\phi$.
