$A=\{0,1,2,3,4,5,6,7,8,9\}$, and $B_1, B_2,... B_{100}$ are arbitrary subsets of $A$.
Prove that there exist at least two subsets $B_i$ and $B_j$ such as $|B_i \Delta B_j|\le2$ without using coding theory
Note: this question is asked in Prove that cardinality of the symmetric difference of subsets less than 3, but the accepted solution is an overkill. The challenge is to revisit the problem and come up with something more elegant
Attempt
I started with proof by contradiction aiming at pigeon hole at some point. I tried to split all subsets of $A$ by number of elements $(0, 1,..., 10)$ and find the max number of subsets from each group. The goal is to show that from all $11$ groups the sum of max numbers of subsets is less than $100$ in this case. But I got stuck
For each $i\in \{1,\dots,100\}$, let $$ S_i=\Big\{B_i,B_i\Delta\{0\},B_i\Delta\{1\},\dots,B_i\Delta\{8\},B_i\Delta\{9\}\Big\} $$ That is, $S_i$ is the set of sets whose symmetric difference with $B_i$ has zero or one elements, or the set of sets which can be obtained from $B_i$ by adding/removing at most one element.
If it were true that $|B_i\Delta B_j|> 2$ for each $i,j$, then the sets $S_i$ would be pairwise disjoint. Indeed, if $C\in S_i\cap S_j$, then you could obtain $B_i$ from $B_j$ by adding/removing at most one element from $B_j$ to get $C$, then adding/removing at most one element from $C$ to get $B_i$, implying $|B_i\Delta B_j|\le 2$.
But you cannot have $100$ pairwise disjoint subsets of the power set of $\{0,1,\dots,9\}$ with $11$ elements each, because $100\cdot 11> 2^{10}$. Contradiction!