Consider which $\bigl{(} A_i \bigr{)}_{i\in I}$ is a chain of subsets of $\mathbb{R}$. We say that a countable chain like $\bigl{(} B_n \bigr{)}_{n\in \mathbb{N}}$ is good if :
- for every $n\in \mathbb{N}$, the set $B_n$ be an element of the main chain $\bigl{(} A_i \bigr{)}_{i\in I}$
- for every $i\in I$ there exists an $n\in \mathbb{N}$ that $A_i \subset B_n $.
Give an example of $\bigl{(} A_i \bigr{)}_{i\in I}$ which for it, does not exist such a good chain $\bigl{(} B_n \bigr{)}_{n\in \mathbb{N}}$.
Such an example surely exists and surely exists of subsets with lebesgue zero measure, because if there does not exist then we can find a maximal zero measure set by using Zorn's lemma.
This has nothing to do with Zorn's lemma. Okay, almost nothing to do with Zorn.
Essentially we say that a chain $\mathcal A$ is good, if there is a countable subchain which is cofinal in $\mathcal A$.
So all you need is to show there is a chain which is not of countable cofinality.
For example $\aleph_1$ is regular, assuming "enough choice", and there is always a chain $(A_i)_{i<\omega_1}$, such a chain cannot be good if $\aleph_1$ is regular.
I don't see a natural approach to this using Zorn's lemma as a means for construction such a chain.