Let's fix an arbitrary point $p \in \Delta_n = \{(x_1, ..., x_n) \in \mathbb{R}^n \ : \ \sum_{i=1}^n x_i = 1 \}$
Could you help me prove that every point in a simplex $\Delta_n$ can be written as a convex combination of $p$ and $q \in C^{(p)} = \{ x \in \Delta_n \ : \ \exists i \in \text{supp}(p) : x_i = 0\}$?
It means that there exists $i \in supp(p)$ such that $q_i =0$.
So given any point $x=(x_1, ..., x_n)$ and any $p=(p_1, ..., p_n)$ it needs to be proved that there exists $q \in C^{(p)}$ such that $$x_k = \alpha p_k + (1- \alpha) q_k, \ \ k=1,...,n, \ \ \alpha \in [0,1]$$
I know that $\Delta \setminus C^{(p)}$ is a neighbourhood of $p$, but I doubt it can be useful here.
Could you help me with this?
It took a while to understand what you're actually asking. Am I right in translating the statement to:
Given points $x=(x_1,\ldots,x_n)$ and $p=(p_1,\ldots,p_n)$, there exists a point $q = (q_1,\ldots,q_n)$ such that
1) There exists $1\leq i\leq n$ such that $p_i \neq 0$ and $q_i=0$,
2) There exists $\alpha\in[0,1]$ for which $x=\alpha p + (1-\alpha) q$.
If that is the case, the first step in solving this, is to understand what $\{\alpha p + (1-\alpha) q \mid \alpha \in [0,1] \}$ actually looks like: this is the line segment between $p$ and $q$.
A next realisation that might be useful is what the boundary of $\Delta_n$ looks like: this is a collection $(n-1)$-simplices. If this isn't quite clear, think of a $2$-simplex, an equilateral triangle, and consider its boundary: this is a collection of three line segments of length 1, three $1$-simplices.
To be more precise, the boundary of $\Delta_n$ is $\partial\Delta_n = \{(x_1, ..., x_n) \in \mathbb{R}^n \ \mid \ \sum_{i=1}^n x_i = 1, \exists i: x_i = 0\}$. The requirement for two points $p$ and $q$ that there exists $i$ such that $p_i\neq 0$ and $q_i=0$ thus translates to the statement that there's (at least) one $(n-1)$-simplex in the boundary of $\Delta_n$ that contains $q$, but does not contain $p$.
Combining the above remarks, we can now rephrase the original statement in a purely geometrical fashion:
Given points $x=(x_1,\ldots,x_n)$ and $p=(p_1,\ldots,p_n)$, there exists a point $q = (q_1,\ldots,q_n)$ such that
1G) There exists at least one $(n-1)$-simplex in the boundary of $\Delta_n$ that contains $q$ but does not contain $p$.
2G) The point $x$ lies on the line segment between $p$ and $q$.
Given the points $x$ and $p$ from the statement, we're now in a position to quite easily find a point $q$. First of all, if $x = p$, just do what the hell you like. If $p\neq x$, draw the line segment starting at $p$ through $x$ and 'consider where it leaves $\Delta_n$'. This is the point $q$ and I'm pretty sure you'll be able to figure out the loose ends from the above. If anything isn't clear, try picturing the situation in equilateral triangles and tetrahedra first.
Notice that I use a rather vague description at the end. This is to avoid writing 'find the point where this line hits the boundary', since that would be incorrect: if there's one boundary component that contains both $p$ and $x$, the entire line segment will lie within this boundary component. A more precise formulation would be: "Let $k\leq n$ be the smallest integer such that $p$ and $x$ lie in a common $k$-simplex in the boundary of $\Delta_n$. The point $q$ is the point on the half-line starting from $p$ through $x$ where $q$ lies in an $m$-simplex in the boundary of $\Delta_n$, where $m<k$." Given the formulation of the question, though, I've got the feeling that this is more complicated than required. If you've got some spare time to think about it, it might be worth it. I'd advise to start thinking about a tetrahedron (where the equilateral triangles at the sides are the 2-simplices in the boundary, the edges are the 1-simplices and the vertices the 0-simplices) having $p$ and $x$ in some of these lower-dimensional simplices.