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.
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.