A decreasing transfinite sequence of subsets of a countable set. Let $X$ be a countable set and $(S_\alpha)_{\alpha< \rho}$ is a decreasing transfinite sequence of subsets of $X$ in the sense that
$$
S_\alpha \supset S_\beta
$$
whenever $\alpha<\beta$. Here $\rho$ is some fixed ordinal.

Suppose that $(S_\alpha)_{\alpha< \rho}$ is strictly decreasing, i.e. $S_\alpha \ne S_\beta$ whenever $\alpha<\beta$, how do we show that $\rho$ must be a countable ordinal?

I am sorry if this question is elementary, I have very little training in axiomatic set theory. I think I could prove by contradiction but some crucial steps are missing and I don't know how to make it rigorous.

If it is strictly decreasing, then $A_\alpha=S_\alpha\setminus S_{\alpha+1}$ is non-empty, and $A_\alpha\cap A_\beta=\varnothing$ whenever $\alpha\neq\beta$.
How many pairwise disjoint subsets a countable set can have?
