About Euclidean distances $\newcommand\R{\mathbb R}$Let $0<d_1<\cdots<d_k<\infty$ and let $m_1,\dots,m_k$ be any integers $\ge1$. Let $n:=m_1+\dots+m_k-1$.
Let $d$ denote the Euclidean distance in $\R^n$.

Do then there always exist pairwise disjoint subsets $S_1,\dots,S_k$ of $\R^n$ of respective cardinalities $m_1,\dots,m_k$ such that
$$d(x,y)=d_j$$
for all $j\in\{1,\dots,k\}$ and all $(x,y)\in T_j\times S_j$ with $x\ne y$, where $T_j:=S_1\cup\dots\cup S_j$?

This conjecture is obviously true for $k=1$ and seems not hard to prove for $k=2$.

Reviewing the edits suggested by MattF., one can restate the conjecture in the following way:

Let $n$ be any natural number. Suppose that $0<d_1\le\cdots\le d_{n+1}<\infty$.


Do then there always exist pairwise distinct points $x_1,\dots,x_{n+1}$ in $\R^n$ such that
$$d(x_i,x_j)=d_j$$
for all integers $i,j$ such that $1\le i<j\le n+1$?

 A: Finite ultrametric spaces allow isometric embeddings into $L_2$.
A: [Trying to rewrite everything]
Lemma 1 The altitude of a regular simplex of side $a$ is larger than $a/\sqrt2$.
Proof Folllows from the fact that the central angle of every edge of a regular simplex is obtuse.
Lemma 2 Let $P$ be either (i) $\mathbb R^n$, or (ii) an $n$-dim sphere (in $\mathbb R^{n+1}$). Then there exist $n+2$ points in $P$ such that
$$
  \text{all their pairwise distances equal $a$, except for one which is at least $a$} 
  \qquad (*)
$$
iff either (i) holds, or (ii) holds, with the sidelength of the inscribed $(n+2)$-vertex simplex being at least $a$.
Proof If $P$ does not contain $n+1$ points forming a regular simplex of side length $a$, then both conditions are violated.
Otherwise, consider such simplex. Let $O$ be the center of the sphere, $M$ be the center of the simplex, and $N$ be the centerr of one of its facets, $A$ being the complementary vertex of the simplex. When the radius of the sphere grows, $OM$ grows as well, while $OM\perp MN$ always.
If (i) holds, then $n+2$ points are obtained by reflecting the vertex in the corresponding facet, and by Lemma 1 we are done. Assume (ii). To get the $(n+2)$th point $A'$ on $P$ equidistant from the facet's vertices, we need to reflex $A$ in $ON$; since $\angle ONM$ is growing, the distance $AA'$ is increasing, wo $AA'\geq a$ iff the sphere is large enough in order to fit the regular simplex with side $a$.
$$ $$
Now, ket $P$ be as in Lemma 2, satisfying $(*)$ with $a=d_k$. We claim that the required sets $S_1,\dots, S_k$ can be found on $P$.
Induction on $k$. The base case $k=1$ is provided by Lemma 2.
For the step, choose a set $S$ of $n+2$ points on $P$ satisfying $(*)$ with $a=d_k$. Then $m_k$ of them go to the set $S_k$. Other sets to be constructed should lie on $P$, as well as on the $(n-m_k+1)$-dimensional sphere cpnsisting of points at distance $d_k$ from all points of $S_k$. The intersection of those spheres is an $(n-m_k)$-dimensional sphere containing $n-m_k+2$ points (from $S$) satisfying $(*)$ with $a=d_k$, so a fortiori with $a=d_{k-1}$; hence we can apply the inductive hypothesis to this sphere getting the other desired sets.
