Countably Infinitely Many Points in a Euclidean Space with Restraining Orders

Main Question. An answer is affirmative already for $d=3$. Indeed, for each $n$ put $$x_n=\left(\frac 12\cos\frac{\pi}{n+1}, \frac 12\sin\frac{\pi}{n+1}, 0\right)$$ and $$y_n=\left(-\frac 12\cos\frac{\pi}{n+1}, -\frac 12\sin\frac{\pi}{n+1}, \sin\frac{\pi}{2(n+1)(n+2)+1}\right).$$ Thus $\{x_n\}$ is a sequence of points of a circle in a plane $Oxy$ with radius $\frac 12$ centered at the origin, and $y_n=-x_n+\left(0,0, \sin\frac{\pi}{2(n+1)(n+2)+1}\right)$ for each $n$. It follows $\|x_n-y_n\|_2> 1$.

If $i<j$ then $$4\|x_i-x_j\|^2_2\le 4\|y_i-y_j\|^2_2=$$ $$\left(\cos\frac{\pi}{i+1}-\cos\frac{\pi}{j+1}\right)^2+ \left(\sin\frac{\pi}{i+1}-\sin\frac{\pi}{j+1}\right)^2+ 4\left(\sin\frac{\pi}{2(i+1)(i+2)+1}- \sin\frac{\pi}{2(j+1)(j+2)+1}\right)^2<$$ $$2-2\cos\frac{\pi}{i+1}\cos\frac{\pi}{j+1}-2\sin\frac{\pi}{i+1}\sin\frac{\pi}{j+1}+4\sin^2\frac{\pi}{2(i+1)(i+2)+1}=$$ $$2-2\cos\left(\frac{\pi}{i+1}-\frac{\pi}{j+1}\right)+ 4\sin^2\frac{\pi}{2(i+1)(i+2)+1} =$$ $$4 \sin^2\left(\frac{\pi}{2(i+1)}-\frac{\pi}{2(j+1)}\right)+ 4\sin^2\frac{\pi}{2(i+1)(i+2)+1}<$$ $$4 \sin^2\frac{\pi}{2(i+1)}+ 4\sin^2\frac{\pi}{2(i+1)(i+2)+1}\le $$ $$4 \sin^2\frac{\pi}{4}+ 4\sin^2\frac{\pi}{13}<4 \sin^2\frac{\pi}{4}+ 4\sin^2\frac{\pi}{6}=3<4.$$

If $i\ne j$ then

$$4\|x_i-y_j\|^2_2=$$ $$\left(\cos\frac{\pi}{i+1}+\cos\frac{\pi}{j+1}\right)^2+ \left(\sin\frac{\pi}{i+1}+\sin\frac{\pi}{j+1}\right)^2+ 4\sin^2\frac{\pi}{2(j+1)(j+2)+1}=$$ $$2+2\cos\frac{\pi}{i+1}\cos\frac{\pi}{j+1}+2\sin\frac{\pi}{i+1}\sin\frac{\pi}{j+1}+4\sin^2\frac{\pi}{2(j+1)(j+2)+1}=$$ $$2+2\cos\left(\frac{\pi}{i+1}-\frac{\pi}{j+1}\right)+ 4\sin^2\frac{\pi}{2(j+1)(j+2)+1} =$$ $$4-4\sin^2\left(\frac{\pi}{2(i+1)}-\frac{\pi}{2(j+1)}\right)+4\sin^2\frac{\pi}{2(j+1)(j+2)+1}.$$

So it suffices to show that $$\left|\sin\left(\frac{\pi}{2(i+1)}-\frac{\pi}{2(j+1)}\right)\right|>\sin \frac{\pi}{2(j+1)(j+2)+1}.$$ Put $m=\min\{i,j\}$. Then $$\left|\sin\left(\frac{\pi}{2(i+1)}-\frac{\pi}{2(j+1)}\right)\right|\ge\sin\left(\frac{\pi}{2(m+1)}-\frac{\pi}{2(m+2)}\right)=$$ $$\sin \frac{\pi}{2(m+1)(m+2)}>\sin \frac{\pi}{2(j+1)(j+2)+1}.$$

Related question. I guess by $\ell^2(\Bbb N,\Bbb R)$ you mean the usual $\ell_2$. Then the negative answer for an uncountable $J$ follows from the following

Proposition. If a metric space $(X,\rho)$ has contains families $\{x_i\}$ and $\{y_i\}$ for $i\in J$ satisfying the respective conditions then $|J|\le d(X)$, where $d$ is the density of the space $X$.

Proof. For each $i\in J$ pick $\varepsilon_i>0$ such that $\rho(x_i, y_i)>1+2\varepsilon_i$. Then an open ball $B\left(x_i, 2\varepsilon_i\right)$ of radius $2\varepsilon_i$ centered at $x_i$ contains no other $x_j$. It follows that a family $\left\{B\left(x_i, \varepsilon_i \right):i\in J\right\}$ is a family of mutually disjoint non-empty open subsets of $X$. $\square$