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, $\|u-v\|_2=1$.