Let $d_1(x,y)=|x-y|$ and $d_2(x,y)=|\frac1x-\frac 1y|$. Prove that $d_1$ and $d_2$ are equivalent in $(0,1)$
First proof
Check the validity of the proof please. Im interested in easier or simpler proofs too, if exists.
Two metrics $d_1$ and $d_2$ over some set $X$ are defined as equivalent if and only if
$$(\forall x\in X)(\forall \epsilon >0)(\exists r,s>0):(B_1(x,r)\subseteq B_2(x,\epsilon))\land (B_2(x,s)\subseteq B_1(x,\epsilon))$$
where $B_j(x,\epsilon):=\{y\in X: d_j(x,y)<\epsilon\}$ are open balls.
Part one: for all $x,y\in(0,1)$ and $\epsilon>0$ we have that
$$\begin{align}B_1(x,r)\subseteq B_2(x,\epsilon)&\iff |x-y|<r\implies \left|\frac1x-\frac 1y\right|<\epsilon\\&\iff |x-y|<r\implies|x-y|<xy\epsilon\\&\iff r\le xy\epsilon\end{align}$$
Case 1: if $1>x\epsilon$ and cause $|\frac1x-\frac1y|<\epsilon$ then we have that
$$\left|\frac1x-\frac1y\right|<\epsilon\implies\begin{cases}\frac1y>\frac1x-\epsilon,&\text{if } \frac1y<\frac1x\\\frac1y<\frac1x+\epsilon,&\text{if }\frac1y\ge \frac1x\end{cases}\implies\begin{cases}y<\frac{x}{1-x\epsilon},&\text{if }y>x\\y>\frac{x}{1+x\epsilon},&\text{if }y<x\end{cases}$$
in any case $y$ is bounded below positively because $x>0$, what means that $\inf\{xy:x,y\in(0,1)\}\neq 0$. Then exists some $r>0$ such that $r\le xy\epsilon$, i.e.
$$(\forall x\in (0,1))(\forall\epsilon\in(0,x))(\exists r\in\Bbb R_{>0}):B_1(x,r)\subseteq B_2(x,\epsilon)$$
Case 2: If $1<x\epsilon$ then we can set some $\delta>0$ such that $1>x\delta$, reducing this case to the above case, i.e.
$$B_1(x,r)\subseteq B_2(x,\delta)\subsetneq B_2(x,\epsilon)$$
Part two: for all $x,y\in(0,1)$ and $\epsilon>0$ we have that
$$\begin{align}B_2(x,s)\subseteq B_1(x,\epsilon)&\iff \left|\frac1x-\frac 1y\right|<s\implies |x-y|<\epsilon\\&\iff |x-y|<xys\implies|x-y|<\epsilon\\&\iff xys\le \epsilon\implies s\le\epsilon\end{align}$$
because $\sup\{xy:x,y\in(0,1)\}=1$. Then setting $s=\epsilon$ we have that
$$B_2(x,\epsilon)\subseteq B_1(x,\epsilon),\quad\forall\epsilon>0,\forall x\in(0,1)$$
And with this the proof ends.
Second proof
The open balls of $d_1$ have the form
$$B_1(x,s)=(x-s,x+s)\cap(0,1)$$
The open balls of $d_2$ have the form
$$B_2(x,r)=\left\{y\in(0,1):\left|\frac1x-\frac 1y\right|<r\right\}\implies \frac1y\in\left(\frac1x-r,\frac1x+r\right)$$
then
$$y\in\left(\frac{1-rx}{x},\frac{1+rx}{x}\right)^{-1}=\begin{cases}\left(\frac{x}{1+rx},\frac{x}{1-rx}\right)&\text{if }\frac1x>r\\\left(\frac{x}{1-rx},\infty\right)\cap(0,1)&\text{if }\frac1x<r\end{cases}$$
From here we can check that exists solutions for $B_2(x,r)\subseteq B_1(x,s)$ for any $x\in(0,1)$ and any $s>0$.
And exists solutions too for $B_1(x,r)\subseteq B_2(x,s)$ for any $x\in(0,1)$ and any $s>0$.