Let $\lambda$ be a singular cardinal of countable cofinality.
Is there necessarily a sequence $\{A_\alpha\mid\alpha<\lambda^+\}$ of countable subsets of $\lambda$, such that $\alpha<\beta$ if and only if $A_\alpha\setminus A_\beta$ is finite?
In other words, we know that for $\omega$, there is an uncountable sequence of subsets which is increasing modulo finite changes. Is the same true for other cardinals of countable cofinality?
For $\lambda > 2^{\aleph_0}$, there is no such sequence.
Suppose $\lambda > 2^{\aleph_0}$. Because $2^{\aleph_0}$ cannot have countable cofinality, there is some $\kappa < \lambda$ with $2^{\aleph_0} < \kappa$. Consider the sequence $\{A_\alpha \cap A_\kappa \mid \alpha < \kappa\}$. For each particular $\alpha$, the sets $A_\alpha$ and $A_\alpha \cap A_\kappa$ differ by only finitely many elements. Each set of this form is a countable subset of the countable set $A_\kappa$, so there are fewer than $\kappa$ possibilities for the sets $A_\alpha \cap A_\kappa$. Thus we may find $\alpha < \beta < \kappa$ such that $A_\alpha \cap A_\kappa = A_\beta \cap A_\kappa$. But then $A_\alpha$ and $A_\beta$ differ by only finitely many elements, and in particular we have $A_\beta \setminus A_\alpha$ finite while $\alpha < \beta$, contrary to your condition.
