The following argument does not use multivariate calculus:
For $r\geq0$ put $r':=\arctan r$. Looking at the graph of $\arctan$ one immediately sees that $r'\leq r$ and $|r'-s'|\leq|r-s|$.
It suffices to consider the two-dimensional situation. Given the two points $x=(r,0)\ ,\quad y=s(\cos\theta,\sin\theta)$ with $r\geq0$, $s\geq0$ we consider the "scaled" points $x':=(r',0)\ ,\qquad y':=s'(\cos\theta,\sin\theta)\ .$ In terms of $x'$, $y'$ your "new distance" $\delta(x,y)$ satisfies $\eqalign{\Bigl({\pi\over2}\delta(x,y)\Bigr)^2=|x'-y'|^2&=r'^2+s'^2-2r's'\cos\theta\cr &=(r'-s')^2+2r's'(1-\cos\theta)\cr &\leq(r-s)^2+2rs(1-\cos\theta)\cr &=|x-y|^2\ .\cr}$ This proves $\delta(x,y)\leq{2\over\pi}|x-y|\ .$ Letting $x=0$, $y\to0$ one easily verifies that this inequality is best possible.