I'm trying to solve Problem II.I.10 in textbook Analysis I by Amann/Escher.
Two metrics $d_{1}$ and $d_{2}$ on a set $X$ are called equivalent if, for each $x \in X$ and $\varepsilon>0$, there are positive numbers $r_{1}$ and $r_{2}$ such that $\mathbb{B}_{1}\left(x, r_{1}\right) \subseteq \mathbb{B}_{2}(x, \varepsilon)$ and $\mathbb{B}_{2}\left(x, r_{2}\right) \subseteq \mathbb{B}_{1}(x, \varepsilon)$.
Could you please verify whether my attempt on (b) and (c) is fine or it contains logical gaps/errors?
My attempt:
To make the proof easier to follow, I denote the natural metric on $\mathbb R$ by $d_1$, and the metric $X \times X \to \mathbb R, \quad (x,y) \mapsto |(1 / x)-(1 / y)|$ by $d_2$.
(b) Given $x \in X$ and $\varepsilon>0$.
First assume $ y \in \mathbb{B}_{1}\left(x, r_{1}\right)$ or equivalently $|y-x| < r_1$. We have $y \in \mathbb{B}_{2}(x, \varepsilon) \iff |(1 / x)-(1 / y)| = |y-x| / |xy| < \varepsilon$. It suffices to choose $r_1$ such that $r_1 < \varepsilon |xy|$ and $r_1 \le \min \{x,1-x\}$. On the other hand, $|y| \ge |x-r_1|$. Our task is done if we choose $r_1$ such that $r_1 < \varepsilon |x| |x-r_1|$.
Next assume $ y \in \mathbb{B}_{2}\left(x, r_{2}\right)$ or equivalently $|(1 / y)-(1 / x)| = |y-x| / |xy| < r_2$. We have $y \in \mathbb{B}_{1}(x, \varepsilon) \iff |y-x| < \varepsilon$. It suffices to choose $r_2$ such that $r_2 |xy| < \varepsilon$ and $r_2 \le \min \{x,1-x\}$. On the other hand, $|y| \le |x+r_2|$. Our task is done if we choose $r_2$ such that $r_2|x||x+r_2|< \varepsilon$.
(c) Assume metric $d_3$ on $\mathbb R$ is equivalent to $d_1$ and induces $d_2$. Take the sequence $(x_n)_{n \in \mathbb N}$ in $X$, where $x_n = 1/(n+2)$. Since $x_n \xrightarrow{d_1} 0$, $x_n \xrightarrow{d_3} 0$ and thus there exists $U \in \mathbb R$ such that $d_3 (x_{n}, 0) \le U$ for all $n$.
We have $d_3(x_n, 0) \ge d_3(x_n, x_{2n}) - d_3 (x_{2n}, 0) = d_2(x_n, x_{2n}) - d_3 (x_{2n}, 0) = n - d_3 (x_{2n}, 0)$ $\ge n - U$. This contradicts $x_n \xrightarrow{d_3} 0$.