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This answer refers to an earlier, slightly inaccurate, version of the problem. (GK)

I think your conditions might be insufficient, even if in (2) you require the intersection to be a convex set. If d=1, first take three intervals, A, B and C. Your sets can be $A\cup B$, $B\cup C$ and $A\cup C$. The intersection of any two will be an interval.

A similar example if d=2 is to take four squares, $A=[0,1]\times [0,1]$, $B=[0,1]\times [1,3]$, $C=[1,3]\times [0,1]$, $D=[1,3]\times [1,3]$. Now take the four sets to be $conv(A,B)\cup C$, $conv(B,C)\cup D$, $conv(C,D)\cup A$, $conv(D,A)\cup B$. The intersection of any three sets will be a square.
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I think your conditions might be insufficient, even if in (2) you require the intersection to be a convex set. If d=1, first take three intervals, A, B and C. Your sets can be $A\cup B$, $B\cup C$ and $A\cup C$. The intersection of any two will be an interval.

A similar example if d=2 is to take four squares, $A=[0,1]\times [0,1]$, $B=[0,1]\times [1,3]$, $C=[1,3]\times [0,1]$, $D=[1,3]\times [1,3]$. Now take the four sets to be $conv(A,B)\cup C$, $conv(B,C)\cup D$, $conv(C,D)\cup A$, $conv(D,A)\cup B$. The intersection of any three sets will be a square.