Show that for the metric $\delta(x,y):=\arctan(|x-y|)$ on $\mathbb{R}$ the open sets are the same as on the usual metric
$d(x,y):=|x-y|$
My attempt:
The set of all open sets on $\mathbb{R}$ in regards of the metric $d(x,y)$ is $U:=\lbrace V\subseteq \mathbb{R}:\forall x \in V\,\,\,\exists \epsilon>0:B_{d_\epsilon}(x)\subset V\rbrace$
with $B_{d_\epsilon}(x):=\lbrace y\in \mathbb{R}:|x-y|<\epsilon\rbrace$
First I want to show that for all $V\in U$ the set on the metric $\delta(x,y)$ is also open:
We choose any $V\in U$, which means $x \in V\Longrightarrow \exists \epsilon>0:B_{d_\epsilon}(x)\subset V$
And we let $B_{\delta_\xi}(x):=\lbrace y \in \mathbb{R}:\arctan(|x-y|)<\xi\le|x-y|\rbrace$
Since $\arctan(|x-y|)<\xi\le |x-y|<\epsilon\Longrightarrow B_{\delta_\xi}(x)\subseteq B_{d_\epsilon}\subset V $
Which proofs, that all open sets in regards of $(\mathbb{R},d)$ are also open in regards of $(\mathbb{R},\delta)$
Now I will show that there doesnt exists a open set on $(\mathbb{R},\delta)$ which isnt open on $(\mathbb{R},d)$:
Let $M$ be a open set in regards to $(\mathbb{R},\delta)$, this implies:
$\forall x \in M\,\,\,\exists \epsilon > 0:B_{\delta_\epsilon}(x)\subset M$
Let now $\eta:=\frac{r}{2}$ with $r:=|\arctan(x-\epsilon)|$. Choosing $\eta$ as our radius in regards of the metric $d$ and the point $x \in M$:
$$B_{d_\eta}(x)\subset B_{\delta_\epsilon}(x)\subset M$$
Which tells us that $M$ is also open in regards of the metric space $(\mathbb{R},d)$
$\Box$
Could someone look over my reasoning and give me feedback? Firstly if its correct, and if not what I need to change. Thank you alot for your time!