Shrinking subset and product

OK, it seems that this is a counterexample.

Take 8 disjoint segments $I_1,\dots,I_8$ of length 1. Take 8 sets $$ S_1=\{1,2,3\}, \quad S_1=\{4,5,6\}, \quad S_1=\{1,2,4\}, \quad S_1=\{1,2,5\}, \quad S_1=\{1,2,6\}, \quad S_1=\{1,3,4\}, \quad S_1=\{1,3,5\}, \quad S_1=\{1,3,6\}. $$ Say that $I_i$ lies in $A_j$ iff $j\in S_i$, otherwise $I_i$ and $A_j$ are disjoint. Finally, set $c=2$.

In this situation, the optimal $B$ is $I_1\cup I_2$. where the product equals $1$. This follows from the fact that $\sum_j |B\cap A_j|\leq 6$, since any point is covered by at most three of the $A_j$.

Now set $A_1’=A_1\setminus I_1$. Consider the quantities $$ x=\left|B’\cap\left(\bigcup_{i=3}^8 I_i\right)\right|, \quad y=|B’\cap I_2|. $$ Then, by AM—GM, $$ |B’\cap A_1’|\leq x, \quad \prod_{j=2}^3 |B’\cap A_j|\leq (x/2+2-x-y)^2, \quad \prod_{j=4}^6 |B’\cap A_j|\leq (x/3+y)^3, $$ and the equalities are achievable simultaneously. Hence, in the optimal case, we have $$ \prod_{j=1}^6|B’\cap A_j|=x(2-x/2-y)^2(x/3+y)^3 =\frac{6x\cdot (36-9x-18y)^2\cdot (4x+12y)^3}{6\cdot 18^2\cdot 12^3}. $$

So we seek for a maximizer $(x_0,y_0)$ of $$ f(x,y)= 6x\cdot (36-9x-18y)^2\cdot (4x+12y)^3 $$ under the conditions $x.y\geq 0$, $x+y\leq 2$. We claim that such a maximizer has $x_0\geq 24/17$, which provides $|B’\cap A_1’|>|B\cap A_1|=1$, as desired.

Indeed, we have $$ f\left(\frac{24}{17},\frac{10}{17}\right) =\frac{144}{17}\cdot \left(\frac{216}{17}\right)^5. $$ On the other hand, if $x\leq 24/17$, by AM—GM we have $$ f(x,y)\leq 6x\cdot\left(\frac{2(36-9x-18y)+3(4x+12y)}5\right)^5 =6x\cdot\left(\frac{72-6x}5\right)^5; $$ the right hand part is an increasing function for $0\leq x\leq 2$, so $$ f(x,y)\leq 6\cdot \frac{24}{17}\cdot\left(\frac{72-6\cdot 24/17}5\right)^5 = f\left(\frac{24}{17},\frac{10}{17}\right), $$ as desired.