TITLE: Proving that the intersection of closed convex sets is nonempty
QUESTION [5 upvotes]: This question comes from Section I Chapter 7 of Barvinok's "A Course in Convexity". The statement is as follows:

Let $A_1,A_2,A_3\subset\mathbb{R}^d$ be closed convex sets such that $A_1\cap A_2\neq\emptyset$, $A_1\cap A_3\neq\emptyset$, $A_2\cap A_3\neq\emptyset$ and $A_1\cup A_2\cup A_3$ is a convex set. Prove that $A_1\cap A_2\cap A_3\neq\emptyset$.

This clearly resembles Helly's theorem, but it's a stronger result. The approach that I'm inclined to take is to somehow use the inclusion-exclusion formula, as this is what is discussed in the text shortly before this problem. Possibly in conjunction with the Euler characteristic:
(Theorem 7.4 in Barvinok) There exists a unique valuation $\chi:\mathcal{C}(\mathbb{R}^d)\rightarrow\mathbb{R}$, called the Euler characteristic, such that $\chi([A])=1$ for every non-empty closed convex set $A\subset\mathbb{R}^d$. Note that the notation $\mathcal{C}(\mathbb{R}^d)$ is used to mean the algebra of closed convex sets (i.e. generated by indicator functions).
Beyond this, I'm not sure where to begin. Any advice is greatly appreciated.

REPLY [2 votes]: Ah! You have the hint! From inclusion-exclusion
$$\chi(A\cup B\cup C)=\chi(A)+\chi(B)+\chi(C)-\chi(A\cap B)-\chi(A\cap C)-\chi(B\cap C)+\chi(A\cap B\cap C).$$
All the sets there, apart from perhaps $A\cap B\cap C$ have Euler characteristic
$1$, since they are closed, non-empty and convex. Therefore
$$1=1+1+1-1-1-1+\chi(A\cap B\cap C).$$
Then $A\cap B\cap C$ has nonzero Euler characteristic, and is non-empty.