prove metric space X isn't isometric to any subspace of $\Bbb E^n$ for any $n$ Let $X={A,B,C,D}$ with $d(A,D)=2$, but all other distances equal to 1. $d$ Is a metric. Prove that metric space $X$ is not isometric to any subset of $\Bbb E^n$, for any $n$. 
I've only managed to prove that it's not an isometry when $n=1$. 
Let $T:X \rightarrow \Bbb E^n$ be an isometry and let $n=1$, so the points must be on a line. Because it's an isometry we know $d(T(A),T(D))=2$, $d(T(A),T(B))=1$,$d(T(B),T(D))=1$, so B is in the middle of line segment AD. If we do this for C aswel, we see that C is also in the middle of line segment of AD, which gives a contradiction. So metric space X isn't isometric with $\Bbb E^1$. 
How can I prove this for any $n$?

Assume the contrary that there is $\Phi : X=\{A, B, C, D\} \to \mathbb E^n$ so that  $d(x, y) = |\Phi(x)- \Phi(y)|$. Call $a = \Phi(A)$ (and similarly for $B, C, D$). 
Note that $a, b, c$ and $b, c, d$ forms two equilateral triangles with side length one and sharing the same side $bc$. So the largest possible distance between $a$, $d$ happens when both triangles lie in the same plane. But even in this case $|a-d|=\sqrt 3$ is shorter than $2$. Thus it is impossible to isometric embed $X$ into any Euclidean spaces. 
