Can every well ordered chain be reduced to a countable chain?

Question

Let $X$ be a set, $I$ a well ordered set and $\{X_i\}_{i\in I}$ a family of subsets of $X$ such that

$$X_i\subseteq X_j\mbox{ for }i\leq j$$ $$\bigcup_{i\in I}X_i=X$$

Does there always exist a countable sequence of indexes $(i_n)\subset I$ such that

$$\bigcup_{n\in\mathbb{N}}X_{i_n}=X$$ ?

Answer 1

NO. Not if you have the axiom of choice. For example when $X=I$ and $X$ is an uncountable regular cardinal, ordered by $\in.$

Answer 2

No there isn't, most of the time.

For instance assume $X$ is a(n uncountable) cardinal with uncountable cofinality, for instance $\omega_1, \aleph_{\omega_1},...$ (and many many others), then the family $(\alpha)_{\alpha < X}$ is a well-ordered family satisfying the conditions you gave, but there is no such countable subfamily, by definition of cofinality