Ham Sandwich Theorem. Given 3 measurable "objects" in $\mathbb{R}^3$, it is possible to divide all of them in half (with respect to their measure, i.e. volume) with a single 2-dimensional plane.
Can we turn the following idea into a constructive proof of the Ham sandwich theorem?
Let $B^3_1$, $B^3_2$ and $B^3_3$ be balls in $\mathbb{R}^3$. It is clear that we can divide every volume in two with a plane that intersects the centers of those balls. But now we are free to morph upper and lower halves in an "isovolumetric" fashion without changing the fact that the plane halves the volume. Also, if there are volume migration between the sides of the plane, then the net change must be zero.
If we divide each half ball into a large, but finite, collection of cubes, then it should be clear that we can approximate any shape with equal volume.
Unfortunately, you are not free to morph the two halves in the very restricted fashion described in your question. You simply do not have that much control over the three given sets. Those three sets can each be much wilder and off kilter than you are imagining.
If your proof were correct, then the given plane $P$ that you found, passing through the centers of the three balls $B^3_1,B^3_2,B^3_3$, would have to work for any three measurable sets $A_1,A_2,A_3$, dividing each of those three sets into two equal volumes. But you can easily see that this is false: simply let $A_1,A_2,A_3$ be obtained, respectively, by translating $B^3_1,B^3_2,B^3_3$ all over to one side of $P$.
Part of the reason that this problem is hard is that a general measurable set $A$ does not have any point like the center of a ball or like the symmetry point of a centrally symmetric object. That is, there does not generally exist a point $P$ having the property that any plane through $P$ cuts $A$ into two subsets having equal volume.