Throughout this question, the variable $I$ represents the set of integers $\{0,\dots,p\}$, with $p > 0$. I'll use the ad hoc notation $I_j$ to denote the set $I \, \backslash \{j\}$, and similarly, $I_{j, k}$ to denote the set $I \, \backslash \{j, k\}$, etc.
Let $\{x_i\}_{i \in I}$ be $p+1$ distinct vectors in some arbitrary vector space. I want to show that if, for some $k \in I$, the set of vectors $\{(x_i - x_k)\}_{i\in I_k}$ is linearly independent, then, for any $l \in I$, the set of vectors $\{(x_i - x_l)\}_{i\in I_l}$ is linearly independent.
This result seems to me intuitively obvious. Basically, the premise of the proposition expresses in terms of linear independence what it means to have $p+1$ points in "general position", and the proposition asserts that this linear independence property does not hinge on the choice of "basepoint".
I can prove the proposition, but my proof strikes me as too complicated and hard-to-follow for such an intuitively obvious result. Since I have an uncanny talent for overcomplicating things, I suspect that there is a far simpler/more immediate proof that I'm somehow not seeing.
FWIW, here's my proof.
If $l = k$ the conclusion holds by assumption. So let $l \neq k$.
Given any linear combination of the vectors $\{(x_i - x_l)\}_{i\in I_l}$ having coefficients $\{\alpha_i\}_{i\in I_l}$, we can define a new coefficient $\alpha_l = -\sum_{i\in I_l} \alpha_i$, and then rewrite the original linear combination of the vectors $\{(x_i - x_l)\}_{i\in I_l}$ as a linear combination of the vectors $\{(x_i - x_k)\}_{i\in I_k}$:
$$ \sum_{i\in I_l} (x_i - x_l)\alpha_i = \sum_{i\in I_{k,l}} (x_i - x_k)\alpha_i - (x_l - x_k)\sum_{i\in I_l} \alpha_i = \sum_{i\in I_k} (x_i - x_k)\alpha_i \;. $$
If $\sum_{i\in I_l} (x_i - x_l)\alpha_i = 0$, then the equality above and the assumed linear independence of the vectors $\{(x_i - x_k)\}_{i\in I_k}$ implies that $\alpha_i = 0,\; \forall i \in I_k$. Therefore,
$$\sum_{i\in I_l} (x_i - x_l)\alpha_i = (x_k - x_l)\alpha_k = 0\,.$$
Furthermore, by assumption, $x_k - x_l \neq 0$, so we can also assert that $\alpha_k = 0$. This proves the implication
$$\sum_{i\in I_l} (x_i - x_l)\alpha_i = 0 \;\; \Longrightarrow \;\; \alpha_i = 0, \forall i\in I_l$$
In other words, the vectors $\{(x_i - x_l)\}_{i\in I_l}$ are linearly independent.
It seems that a (most) natural and simple way to prove the claim is to prove Implications $(1 \Rightarrow 2)$ and
$(2 \Rightarrow 3)$.
There exists an index $k \in I$ such that the set of vectors $\{(x_i - x_k)\}_{i\in I_k}$ is linearly dependent.
There exist coefficients $\{\alpha_i\}_{i\in I}$, not all of which are zero, such that $\sum_{i\in I}\alpha_i = 0$ and $\sum_{i\in I} x_i \alpha_i = 0$.
For every index $k \in I$ the set of vectors $\{(x_i - x_k)\}_{i\in I_k}$ is linearly dependent.