A curious generalization of Helly's theorem Here is a curious conjectural extension of Helly's theorem.
It may follow (if true) from a useful theorem of the kind asked in this MO question:

Conjecture: Let ${\cal F}=P_1,P_2,\dots,P_m$ be a family all
  whose members are disjoint union of
  two convex sets in $R^d$. Suppose also
  that 
(1) $m \ge d+2$
(2) Every intersection of $i$
  members of $\cal F$, $i < m$ is also
  the disjoint union of two NONEMPTY compact convex sets.
Then the intersections of all members
  of $\cal F$ is not empty.

Remark: Micha A. Perles showed (in the 70s) that even when $d=2$ you cannot replace "two" by "48".
 A: This answer provides a nice background to the question.
I can say that a theorem similar to the one in the OP is certainly true. The following was conjectured by Grünbaum and Motzkin in [1] and later proved by Amenta in [2].

Theorem. Let $\mathcal C$ be a family of sets in $\mathbb R^d$ such that the intersection of any non-empty finite subfamily of $\mathcal C$ is the disjoint union of at most $k$ closed convex sets. Then the intersection of all sets in $\mathcal C$ is non-empty if and only if the intersection of any $k(d+1)$ elements of $\mathcal C$ is non-empty.

In particular this implies that the conjecture in the OP is true when $m\geq 2d+3$. I believe there are counter examples for $m\le k(d+1)$, showing it is best possible.
References
[1] Branko Grünbaum, Theodore S. Motzkin, "On components in some families of sets" Proceedings of the American Mathematical Society, Vol. 12, No. 4, 607-613, doi:10.2307/2034254, MR0159262, Zbl 0106.01001.
[2] Nina Amenta, "Helly-type theorems and Generalized Linear Programming" "Discrete & Computational Geometry 12, 241–261 (1994), EuDML: 131330, MR1298910, Zbl 0819.90056.
A: 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.
A: This answer shows that you cannot strengthen the conditions of the questions and demand condition (2) only for $m-1$ sets.
Answer for new version IF we only consider the intersection of m-1 sets (as construction fails otherwise, as pointed out by Gil in the comment):
Now it is not hard to prove that the statement is true for d=1 but there is a counterexample for d=2. Take a circle and divide its perimeter into six equal parts, A, B, .., F such that e.g. A, B and C are on the top. Now take a point somewhere high, P, and another somewhere low, Q. Our four sets will be the following: conv(A,P) $\cup$ conv(D,Q), conv(B,P) $\cup$ conv(E,Q), conv(C,P) $\cup$ conv(F,Q) and finally the last set is the disc. Now the intersection of the first three will be around P and Q, while the intersection of the fourth with any two other sets will be around two disjoint arcs of the perimeter.
