Let $f: \mathbb{R} \to \mathbb{R}$ an strictly increasing function. Then $d(x,y)=|f(x)-f(y)|$ is a metric, no problem proving this.Then the topologt generated $d$ and the topology generated by euclidean metric $d_{E}$ are equivalent. To prove this last statement I need to prove the following two things:
(1) For every $x \in \mathbb{R}$ and each $\epsilon > 0$ there is some $\delta > 0$ such we have the contention $B_{d}(x, \delta) \subseteq B_{d_{E}}(x, \epsilon)$.
(2) For every $x \in \mathbb{R}$ and each $\epsilon>0$ there is some $\delta >0$ such we have the contention $B_{d_{E}}(x, \delta) \subseteq B_{d}(x, \epsilon)$.
I prove (2) taking $y \in B_{d_{E}}(x, \delta)$, then $d_{E}(x,y)=|x-y| < \delta$ and by the continuity of $f$ we have $d(x,y)=|f(x)-f(y)| < \epsilon$. This last inequality means $y \in B_{d}(x, \epsilon)$. So $B_{d_{E}}(x, \delta) \subseteq B_{d}(x, \epsilon)$.
Proving (1) seems a lot harder or at least since I dont see how to proceed, I have been thinking to prove $f^{-1}$ exist and this inverse function is continuous but Im not sure. Any help to finish this proof will be apreciated.
There's a catch here: a strictly increasing function need not be continuous. For example, $f(x)=\begin{cases}x+1&x>0\\0&x=0\\x-1&x<0\end{cases}$ is not continuous at zero, and the resulting metric $d(x,y)=f(|x-y|)$ does not induce the usual topology. The ball of radius $\frac12$ centered at zero contains only the point $0$, so the single-element set $\{0\}$ is open in this topology (and, of course, not in the usual topology).
For this to be true, we need the additional hypothesis that $f$ is continuous.
Yes, showing that $f^{-1}$ exists and is continuous (for continuous $f$) is a good way to go. The Intermediate Value Theorem may be helpful here.