Prove arctan metric is equivalent to euclidean metric

Question

Given two metrics in $\mathbb{R}$, $d_1(x,y)=|\arctan x-\arctan y|$, and $d_2(x,y)=|x-y|$.

I need to prove those two metrics are equivalent. I have the definition that two metrics are said to be equivalent if they generate the same open sets.

I suppose that I have to show double inclusion of two open balls: $B_{d1}(x,r_1)\subset B_d2(x,r_2)$ and $B_{d2}(x,r_2)\subset B_{d1}(x, r_1)$ to get $B_{d1}(x,r_1)=B_{d2}(x,r_2)$

Here is my attempt after read GReyes's comment.

let $a\in\mathbb{R}$ and $r>0$, we have a $d_2$-open ball $B_{d1}(a,r)$. $\forall b\in B_{d2}$, $d_2(a,b)<r$. By Mean Value Theorem, we have $|\arctan a-\arctan b|\leq|f^{'}(x)||a-y|$ where $f^{'}x=\frac{1}{1+x^2}$. Hence we have $|\arctan a-\arctan b|\leq|a-b|<r$. So we can find an open-ball $B_{d1}(a,r^{'})$ such that $r^{'}\leq r$ and $B_{d1}(a,r^{'})\subset B_{d2}(a,r)$.