The problem that I'm working one is the following
Let $S$ be a finite set of points in $\mathbb{R}^{n}$. A stamp set $S$ is a set such that every point in $\mathbb{R}^{n} \setminus S$ is an irrational distance from some point in $S$.
A minimal stamp set is a stamp set which has less elements than any other stamp set, and obviously these exist due to the well-ordering principle.
And for some notation let $m(\mathbb{R}^{n})$ denote the order of a minimal stamp set
My conjecture is that $m(\mathbb{R}^{n}) = n+1$ via the stamp set $\{0,{T}_{1}, ... , {T}_{n} \}$ where $0$ is the zero vector and ${T}_{i}$ is the vector with $0$'s everywhere except the $i$-th position, where it contains $T$ a transcendental number.
I currently have no idea how to prove this conjecture, but the statement $m(\mathbb{R}^{n}) = n+1$ is equivalent to the fact that $m(\mathbb{R}^{n}) < m(\mathbb{R}^{n+1})$.
And then again I have no idea how to prove that either
I have a feeling that there is some sort of proof involving analysis due to the fact that one can easily generalize my definition of stamp sets to general metric spaces and easily prove that if there is a bijective isometry between $2$ metric spaces then they have the same order of minimal stamp set.
And along with my conjecture you can prove that there is no bijective isometry from $\mathbb{R}^{n}$ to $\mathbb{R}^{k}$ for $n \neq k$. Which to my knowledge is a non-trivial fact to prove.
I think your conjecture is correct, but I am far from an expert so please excuse any errors or misunderstandings.
Your conjectured stamp set is Sufficient: Consider point $X = (x_1, x_2, \dots, x_n)$ and assume for later contradiction that it has rational distance to all the points in your stamp set. This means:
$$ x_1^2 + \dots + x_n^2 = q_0 \in \mathbb{Q}$$
$$ \forall i: x_1^2 + \dots + x_{i-1}^2 + (T - x_i)^2 + x_{i+1}^2 + \dots + x_n^2 = q_i \in \mathbb{Q}$$
Subtracting, we get:
$$\forall i: (T - x_i)^2 - x_i^2 = T^2 - 2 x_i T = q_i - q_0 \implies x_i = \frac{q_i - q_0 -T^2}{-2T}$$
Substituting back into the first equation, we have:
$$\sum_{i=1}^n (q_i - q_0 - T^2)^2 = 4 T^2 q_0$$
But this is a polynomial in $T$ with rational coefficients, which is impossible since $T$ is transendental. $\square$
$n+1$ points are Necessary: The (hand-wavy?) argument is geometric. Consider $n=3$, and around each point draw a sufficiently large sphere ($2$-sphere) of rational radius. Two such spheres will intersect in a circle ($1$-sphere), and the third sphere will intersect it at two points ($0$-sphere). These points have rational distances to all $n=3$ original points. The same argument generalizes to higher $n$.