Proving $d_2(x,y)=|\arctan x-\arctan y|$ is topologically equivalent to the absolute value metric on $\mathbb{R}$ using containment. It seems to be standard in a first course of topology to prove that the metric standard absolute value metric $d_1$ and the metric $d_2(x,y)=|\arctan x-\arctan y|$ are topologically equivalent on $\mathbb{R}$. I have looked at this answer but I do not really understand why this notion of "pullback" allows us to conclude so simply. The only way I am currently comfortable showing the equivalence of two topologies is showing equivalence of the bases. In this case this is showing that for any $x\in \mathbb{R}$ and $r>0$ there exists a $r',r''>0$ such that $B_{d_2}(x,r')\subseteq B_{d_1}(x,r)$ and $B_{d_1}(x,r'')\subseteq B_{d_2}(x,r)$.
Now using the MVT it follows that $|\arctan x- \arctan y|\leq |x-y|$ for all $y\in \mathbb{R}$ and so the second containment follows simply enough. The first containment, however, is giving me a headache. It is quite simple applying the MVT on $[x-r,x+r]$ to get some constant $k$ such that for all $y$ in the open interval $I=(x-r,x+r)$ we have
 $$|x-y|\leq (1+k^2)|\arctan x- \arctan y|.$$
It is tempting to just  choose $r'=\frac{r}{1+k^2}$, but this is where the issue bites. We now pick an arbitrary $y\in B_{d_2}(x,r')$, and if we use this choice it is then very attempting to just use the above inequality and go home. Unfortunately, however, we can only use the above inequality if $y\in I$, which is exactly what we're trying to prove. We cannot use any other intervals that I can think of, and so I think this MVT approach is flawed. I would really like prove equivalence using this containment method, but I am out of ideas. Any help would be much appreciated at this stage.
EDIT: I believe I have successfully answered my question below (If I haven't please let me know), but I'm not going to accept it just yet because I'm interested if someone can come up with a more explicit construction of $r'$.
 A: This question has been preventing me from falling asleep, and in my half awake state I believe my subconscious has come up with an answer. We know that $\tan:(-\pi/2;\pi/2)\to\mathbb{R}$ is continuous, and that for any $y\in \mathbb{R}$ $\arctan y\in(-\pi/2;\pi/2)$. In particular $\tan$ is continuous at $\arctan x$. Thus for $r>0$ there exists an $r'>0$ such that
\begin{align*}
|\arctan x-\arctan y|<r' & \implies |\tan(\arctan x) -\tan(\arctan y)|<r \\
& \iff|x-y|<r
\end{align*}
Hence there does exist an $r'>0$ such that $B_{d_2}(x,r')\subseteq B_{d_1}(x,r)$. I believe this is what the linked post was getting at but I was just too slow to see it at the time. If anyone has an explicit way to construct the $r'$ I'd still very much like to see it, otherwise I think this will suffice.
A: So, let $x$ be a real number; we want a positive real number $r'$ such that $B_{d_2}(x,r')\subseteq B_{d_1}(x,r)$. 


*

*Suppose we have such an $r'$.


Let $y$ lie in $B_{d_2}(x,r)$. Thus we have $y\in B_{d_1}(x,r)$ and $|x-y|<r$.
Define $x'=\arctan x$ and $y'=\arctan y$. Then, we have $x=\tan x'$ and $y=\tan y'$ with $x',y'$ in the interval $(- \frac\pi 2, \frac\pi 2)$. Since $\tan$ is continuously differentiable there, we have
$$ |x-y| = |\tan x' -\tan y'| \leq 
\sup_{t\in[x',y']} (1 + \tan^2 (t))\, |x'-y'| = \sup_{z\in[x,y]}(1+z^2)\, |x'-y'| $$
the second equality holds because $\tan$ is an increasing bijection from $[x',y']$ onto $[x,y]$. Moreover $[x,y]\subseteq[x-r,x+r]$. Thus,
$$ |x-y| < (1+(x+r)^2)\,r' $$


*

*Now, we see that if $r'>0$ is such that $(1+(x+r)^2)\,r' \leq r$, then we're all good.

