Is any finite collection of points contained in a cut and project set with $\mathbb{R}^d$ as internal space? In "Meyer Sets and their Duals" Moody proves that any Meyer set union a finite number of points is again a Meyer set. Additionally, any Meyer set is contained in a finite union of model sets whose internal space (the space you are projecting from) is $\mathbb{R}^d.$ Thus it is clear that any finite collection of points in $\mathbb{R}^m$ is contained in many model sets. 
Given a finite set $\Lambda \subset \mathbb{R}^m$ can we find a $\textbf{single}$ cut and project set containing $\Lambda$ with internal space $\mathbb{R}^d?$ 
 A: The answer is no. 
There are two big issues here. The first is the following: If $\Lambda$ is a subset of a model set $\Lambda(W)$, then it is a subset of the model sets $\Lambda(K)$ for all $W \subset K$. So it is always trivial to construct bigger sets.
Now, if by an unique cut and project set you mean an unique cut and project scheme but potentially multiple model sets, the answer is still no. It is very easy to construct counterexamples, but I will give you instead a more theoretical reasoning why this cannot work.
Assume by contradiction that the claim would be true. Pick a finite set $\Lambda$, and let $(\mathbb R^d \times \mathbb R^n, L)$ be the unique cut and project scheme whose model sets contain $\Lambda$. Pick some $x \notin \pi_1(L)$. 
Then $\Lambda \cup \{ x \}$ is a finite set, thus comes from a cut and project scheme. But since $x$ is not in the projection of $(\mathbb R^d \times \mathbb R^n, L)$,  $\Lambda \cup \{ x \}$ cannot come from the same cut and project scheme. So, there is a different cut and project scheme which produces sets containing $\Lambda \cup \{ x \}$ hence $\Lambda$.
If you want to actually construct them, here is a simple construction. Pick two incommensurable Meyer sets $\Gamma_1$ and $\Gamma_2$, both containing 0. By incommensurable, I mean $\Gamma_1 \cup \Gamma_2$ is not a Meyer set. For example $\Gamma_1=\mathbb Z$, and $\Gamma_2$ the Fibonacci model set or even simpler $\sqrt{2} \mathbb Z$. 
Then, $\Gamma_1 +\Lambda$ and $\Gamma_2 +\Lambda$ are Meyer set, thus by the result you mentioned they come from cut and project schemes, and the incomprehensibility implies that the cut and project schemes are different. But both contain $\Lambda$.
Last but not least, be careful with this. We typically allow for the second group to be a locally compact Abelian group, if you want the internal group to be $\mathbb R^n$ you'll run in some issues with translations. In particular, if I recall right, in this case a Meyer set is not always a subset of a model set, you can only embed it into finitely many translates of one [ this issue doesn't appear if you allow $H$ be any locally compact abelian group]. And to make matters worse, if $\Lambda$ is finite and $S \neq \emptyset$ is any set, $\Lambda$ can be covered by finitely many translates of $S$.
So if you ask the question with finitely many translates, the answer is that actually ANY model set and any cut and project scheme works...
Added If you ask if any finite set in $R^n$ is the subset of a model set with internal space $R^m$, I believe the answer is yes.
Here is how you can probably prove it:
Let $\Lambda$ be a finite set in $R^n$. Define $L$ to be the additive group generated by $\Lambda$. Then, $L$ is a finitely generated subgroup of $R^n$, thus a free group. 
Write 
$$L:= \bigoplus_{i=1}^k \mathbb Z \omega_i \,.$$
Note that by eventually increasing $\Lambda$, you can make $k$ as large as you want. You might also need to assume that $L$ is dense in $R^n$.
Now, all you have to do is prove the following:
Lemma Let $\omega_1,.., \omega_k \in R^n$ be linearly independent over $\mathbb Z$, with $k>n$. Then, there exists $\varphi_1,.., \varphi_k \in \mathbb R^{k-n}$ such that $(\omega_1, \varphi_1),..., (\omega_k, \varphi_k)$ is a basis in $R^k$ and $\sum Z \varphi_k$ is dense in $R^{n-k}$.
And keep in mind during the proof that, if needed, you can make the set $\omega_1,..,\omega_k$ larger by adding some vectors, as long as you don't break the l.i. over Z.
If you prove this lemma, then $(R^n \times R^{k-n}, \bigoplus_{i=1}^k \mathbb Z (\omega_i, \varphi_i))$ is your cut and project scheme.  
