$\newcommand\om\omega\newcommand\Om\Omega\newcommand\de\delta$As in the linked answer, this problem on sets can be restated as the following problem on real numbers.

Let $\Om:=\{0,1\}^n$. For each $\om\in\Om$ and each
$i\in[n]:=\{1,\dots,n\}$, let $\om_i$ denote the $i$th coordinate of
the vector $\om$, so that $\om=(\om_1,\dots,\om_n)$.

For any $a=(a^\om)\in[0,\infty)^\Om$ and any $c\in[0,\sum_{\om\in\Om}
a^\om]$, let
$$B(a):=B_c(a):=\Big\{b=(b^\om)\in[0,\infty)^\Om\colon0\le b^\om\le
a^\om\ \forall\om\in\Om,\sum_{\om\in\Om\setminus\{(0,\dots,0)\}}b^\om=c\Big\}.$$

Let $b(a)=b_c(a)$ be any maximizer of
$$\pi(b):=\prod_{i=1}^n\sum_{\om\in\Om\colon\,\om_i=1}b^\om$$ over all
$b\in B(a)$.

Take now any $u=(u^\om)\in[0,\infty)^\Om$ such that $u^\om\le a^\om$
for all $\om\in\Om$ with $\om_1=1$ and $u^\om=a^\om$ for all
$\om\in\Om$ with $\om_1=0$. Let $b(u)=b_c(u)$ be any maximizer of
$\pi(b)$ over all $b\in B(u)$. Does it then necessarily follow that
$$\sum_{\om\in\Om\colon\,\om_i=1}b(u)^\om\le\sum_{\om\in\Om\colon\,\om_i=1}b(a)^\om?$$

In this restatement, for all $\om\in\Om$

- $a^\om$ stands for $|A^\om|$, where $A^\om:=A^{\om_1}\cap\cdots\cap A^{\om_n}$ and, for $\de\in\{0,1\}$ and $A\subseteq J$, where $J$ is the segment in question, we let $A^\de:=A$ if $\de=1$ and $A^\de:=J\setminus A$ if $\de=0$;
- $b^\om$ stands for $|B\cap A^\om|$.

We see that even the restated problem, in terms of real numbers, concerns maximizing the non-convex/non-concave function $\pi(\cdot)$ over a polytope of dimension $2^n-1$, defined by $2^{n+1}$ affine inequalities. Moreover, we need to compare solutions of two such problems. This seems hard! If someone can answer this question, it could be quite an instructive moment!