In several lists of exercises they affirm that these norms are equivalent: show that this metrics $d_1(s, t)= \left| {s\over 1 +|s|}- {t\over 1 +|t|}\right| $ and $d_2(s,t)=|s-t|$, with $t,s \in \mathbb{R}$, are equivalents.
I tried to show that equivalence of metrics using this definition: for each $ s \in X$, there exist positive constants $\alpha$ and $\beta$ such that, for every point $ t \in X $, $$\alpha d_{2}(s, t) \leq d_{1}(s, t) \leq \beta d_{2}(s, t)$$.
Taking $X=\mathbb{R}$. I choose for example, $s=4$, if that equivalence of norms was truth must exists $\alpha$ and $\beta$ such that, for every $t \in \mathbb{R}$.
$$\alpha |4-t| \leq \left| {4\over 5}- {t\over 1 +|t|}\right| \leq \beta |4-t|$$.
also if $t\geqslant 0$, then $$\alpha |4-t| \leq \left| {4\over 5}- {t\over 1 +t}\right| \leq \beta |4-t|$$. $$\alpha |4-t| \leq {|4-t|\over 5(1 +t)} \leq \beta |4-t|$$ $$5\alpha \leq {1\over (1 +t)} \leq 5 \beta $$ In this part, I can't find an $\alpha>0$ that satisfies, for all $t$ very big in $\mathbb{R}$. Can I conclude that these metrics aren't equivalents. or how can I show that they are equivalents?
It is not possible to have inequalities of the type $$\alpha d_{2}(s, t) \leq d_{1}(s, t) \leq \beta d_{2}(s, t)$$ simply because $d_1(s,t) \leq 1+1=2$ for all $s,t$ and $$\alpha |s-t| \leq d_{1}(s, t) \leq 2$$ cannot holds for all $s,t$.
Your definition of equivalent metrics is called strong equivalence and the two metrics you have are not equivalent in this sense.
Two metrics are (toplogically) equivalent if they have the same open sets. [This is different from strong equivalence]. According to this definition your metrics are indeed equivalent and this has been proved many times on this site. I will let you search MSE for a proof.