I think I have to take two points in $\displaystyle X$ and show that any mapping from $\displaystyle X$ to $\displaystyle \mathbb{E}^n$ would result in a different distance between them.Let $\displaystyle X=\{A,\ B,\ C, \ D\}$ with $\displaystyle d(A,D)=2$, but all other distances equal to 1.

d is a metric.

Prove that the metric space X is not isometric to any subset of $\displaystyle \mathbb{E}^n$ for any n.

Hint: can you realise $\displaystyle X$ as a subset of a sphere $\displaystyle S^2$ of appropriate radius, with the spherical "great circle" metric?

Here's my attempt:

Take a sphere of radius 1 and the points A and D.

Since $\displaystyle d(A,D)=2$, A and D are on opposite sides of the sphere.

Suppose there exists an isometry $\displaystyle \phi:X \rightarrow \sigma$ where $\displaystyle \sigma$ is is the plane used in stereographic projection (also a subset of $\displaystyle \mathbb{E}^n$).

Then $\displaystyle \phi(A)=\phi(D)$. Hence $\displaystyle d(\phi(A),\phi(D))=0$ which is not equal to $\displaystyle d(A,D)=2$.

Hence $\displaystyle \phi$ cannot be an isometry.