Maximum number of elements of $X \subset \mathbb{R}^n$ such that the discrete metric in $X$ is induced by the eucliean one? What is the maximum number of elements of $X \subset \mathbb{R}^n$ such that the discrete metric in $X$ is induced by the eucliean one in $\mathbb{R}^n$?
It's easy to verify that if $n=1$, then the answer is $2$. The case $n = 2$ has already been solved (What is the maximum number of points that a subspace $X \subset \mathbb{R}²$ can have so that $\mathbb{R}²$ induces the discrete metric in $X$?), whose answer is 3. In similar way with solid geomtry, one can prove that if $n=3$, then the answer is 4. So, a reasonable conjecture is $n+1$. However, I couldn't find a solution for $n>3$.
 A: Here is an answer using an isomorphism and a previously derived result about the dimension of $K_n$, the complete graph on $n$ vertices.
Definition: The dimension of a graph $G$, denoted $\dim G$, is the smallest natural number $n$ such that $G$ can be embedded into $\mathbb{R}^n$ with every edge of $G$ having length $1$. Such an embedding is called a unit-embedding.
On page four of author Ryan Kavanagh's paper, Explorations on the Dimension of a Graph, he proves that $\dim K_n = n - 1$ (for more details and examples, see On the Dimension of a Graph here).
Proposition: The maximum number of elements of a subset $X$ of $\mathbb{R}^n$ such that the discrete metric is induced by the Euclidean metric is $n+1$ (here, discrete metric means that any two distinct points in $X$ are a distance one from each other).
Proof: First note that if $X\subseteq\mathbb{R}^n$, then $|X|$ can be $n+1$ by considering the vertices of a regular $n$-simplex with common edge length one.
Suppose $X\subseteq\mathbb{R}^n$ is a set such that the induced metric on $X$ is the discrete metric. Assume for contradiction that $X$ has more than $n+1$ points. Consider a subset $X'\subseteq X$ such that $|X'|=n+2$. Construct the graph $G=(X',E_{X'})$ where $X'$ is the set of vertices of $G$ and $E_{X'}$ is the set of edges of $G$, where an edge is drawn between every pair of points in $X'$. Then $G$ is isomorphic to $K_{n+2}$, since we have a graph on $n+2$ vertices with every pair of vertices connected by an edge. More explicitly, there is an edge preserving bijection $f:V(K_{n+2})\to X'$. Since each edge in $G$ has unit length, then the isomorphism $f$ from $V(K_{n+2})$ to $X'$ is actually a unit-embedding of $K_{n+2}$ in $\mathbb{R}^n$. But this is impossible, since $\dim K_{n+2}=n+1$, which completes the proof.
