Countably Infinitely Many Points in a Euclidean Space with Restraining Orders Update.  The answer to the main question for $d\geq 3$ is positive.  For $d=1$, it is easy to show that the answer is negative.  What is the answer for $d=2$?

Main Question. Let $\mathbb{N}$ be the set of positive integers.  Do there exist $d\in\mathbb{N}$ such that there are pairwise distinct
points $x_1$, $y_1$, $x_2$, $y_2$, $\ldots$ in $\mathbb{R}^d$ such
that
(i) $\left\|x_i-y_i\right\|_2 >1$ for every $i\in\mathbb{N}$,
(ii) $\left\|x_i-y_j\right\|_2<1$ for all $i,j\in\mathbb{N}$ such that
$i\neq j$,
(ii) $\left\|x_i-x_j\right\|_2<1$ and $\left\|y_i-y_j\right\|_2<1$ for all $i,j\in\mathbb{N}$?
Here, $\|\bullet\|_2$ is the usual Euclidean norm on $\mathbb{R}^d$.

Alternative Interpretation. Let $a_1$, $a_2$, $a_3$, $\ldots$ and $b_1$, $b_2$, $b_3$, $\ldots$ be two countably infinite collections of people.  Suppose that $a_i$ and $b_i$ have a restraining order against one another for every $i=1,2,3,\ldots$.  No other pair of people has a restraining order.  However, each person is friends with every other persons except the one against whom they have a restraining order.  For a fixed dimension $d\in\mathbb{N}$, is it possible to arrange these people in the $d$-dimensional Euclidean space so that the distance between each pair of friends is less than $1$, but the distance between each pair with a restraining order is greater than $1$?
This question is based on Graph theory-related problem, unit distance graph, pairs of people with restraining orders.  If we require finitely many points, the answer is positive.  If we require uncountably many points, the answer is negative.  Hence, I wonder what happens if there are countably infinitely many points.

Related Question: Let $J$ be an index set.  For a given cardinality of $J$, do there exist pairwise distinct
points $x_i$ and $y_i$ for $i\in J$ in the real Hilbert space $\ell^2(\mathbb{N},\mathbb{R})$ such
that
(i) $\left\|x_i-y_i\right\|_2 >1$ for every $i\in J$,
(ii) $\left\|x_i-y_j\right\|_2<1$ for all $i,j\in J$ such that
$i\neq j$,
(ii) $\left\|x_i-x_j\right\|_2<1$ and $\left\|y_i-y_j\right\|_2<1$ for all $i,j\in J$?
Here, $\|\bullet\|_2$ is the usual $2$-norm on $\ell^2(\mathbb{N},\mathbb{R})$.


Attempt at the Main Question.
Suppose that $S:=\big\{x_1,y_1,x_2,y_2,\ldots\}\subseteq\mathbb{R}^d$ satisfies the required properties above.  We observe that $S$ is a bounded set, whence, due to the Heine-Borel Theorem, its topological closure $\bar{S}$ is compact, making it also sequentially compact.  Therefore, the sequence $\left\{x_i\right\}_{i\in\mathbb{N}}$ has a subsequence $\left\{x_{i_j}\right\}_{j\in\mathbb{N}}$ converging to a point $u$ in $\bar{S}$.  Now, the sequence $\left\{y_{i_j}\right\}_{j\in\mathbb{N}}$ has a subsequence $\left\{y_{i_{j_k}}\right\}_{k\in\mathbb{N}}$ converging to a point $v\in\bar{S}$.  We denote $u_k$ for $x_{i_{j_k}}$ and $v_k$ for $y_{i_{j_k}}$ for all $k\in\mathbb{N}$.  Hence, $\displaystyle\lim_{k\to\infty}\,u_k=u$ and $\displaystyle\lim_{k\to\infty}\,v_k=v$.
Note that $\left\|u_i-v_i\right\|_2 >1$ and $\left\|u_i-v_j\right\|_2<1$ for all $i,j\in\mathbb{N}$ such that $i \neq j$.  Thus, $$\left\|u-v_j\right\|_2=\displaystyle \lim_{k\to\infty}\,\left\|u_k-v_j\right\|_2\leq 1 \text{ and }\left\|u_i-v\right\|=\displaystyle\lim_{k\to\infty}\,\left\|u_i-v_k\right\|_2\leq 1$$ for all $i,j \in \mathbb{N}$. Ergo, $\|u-v\|_2=\displaystyle\lim_{j\to\infty}\,\left\|u-v_j\right\|_2\leq 1$.  However, $$\|u-v\|_2=\displaystyle\lim_{i\to\infty}\,\left\|u_i-v_i\right\|_2\geq 1\,.$$  That is, $\|u-v\|_2=1$.
 A: 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 $X$. $\square$
A: Regarding the $\ell^2$ version, here is an easy argument you can find these points for $J$ countable, even on a sphere.
Note that if $(e_n)_n$ is the standard basis, then $d(\pm re_n,\pm re_m)=r^2\sqrt 2$ for $n\neq m$, and $d(re_n,-re_n)=2r$. So if we take any $r\in (\frac{1}{2},\frac{1}{\sqrt 2})$, then the sequences $x_n=re_n$, $y_n=-re_n$ are as desired.
An analogous argument works in any $\ell^p$ with $p>1$, and in fact shows that for a given cardinal $\kappa$, you can find the sequences with $\lvert J\rvert=\kappa$ in any $\ell^p(\kappa)$ with $p>1$. Paired with Alex's remark that the cardinality of this $J$ is at most the density of the metric space, it follows that the bound is sharp.
