This is the question that I should have asked before asking this [older question](https://mathoverflow.net/questions/317191/spatial-dimension-of-a-finite-graph).


If $(X,d)$ is a [metric space](https://en.wikipedia.org/wiki/Metric_space), we associate with it a simple, undirected graph, called its *proximity graph* $G(X,d)$ given by $V(G(X,d)) = X$ and $$E(G(X,d)) = \big\{\{x,y\}:x\neq y\in X \text{ and } d(x,y)\leq 1\big\}.$$


As MO user @YCor pointed out in a comment to a recent deleted question, given any (not necessarily finite) simple, undirected graph $G=(V,E)$, the map $d:V\times V\to \mathbb{R}$ given by $d(v,v) = 0$ for $v\in V$, $d(v,w)=1$ iff $\{v,w\}\in E$ and $d(v,w) = 2$ otherwise gives a metric on $V$ such that $G\cong G(V,d)$.


**Question.** If $G=(V,E)$ is a finite graph, is there a positive integer $n\in\mathbb{N}$ and a finite subset $S\subseteq \mathbb{R}^n$ such that $G \cong G(S,||\cdot||),$ where $||\cdot||$ denotes the Euclidean metric that $S$ inherits from $\mathbb{R}^n$?



Is every finite graph isomorphic to the proximity graph of some $S\subseteq \mathbb{R}^n$?
Yes. If $n=|V|$, then for small $\varepsilon$ any metric spaces on $n$ points with distances belonging to $\{1-\varepsilon, 1+\varepsilon\} $ is embeddable to $\mathbb{R}^{n-1} $.


