Suppose $X$ is a topological space, $\{U_i|i\in I\}$ is a family of open subsets of $X$, which is totally ordered: that means for any $i,j\in I$, either $U_i\subset U_j$ or $U_j\subset U_i$. Denote by $U=\bigcup_{i\in I} U_i$.
The question is:
Could we find a countable subset $J$ of $I$, such that $U=\bigcup_{j\in J}U_j$ ?
This is apparently true for $X$ with countable topological basis. What about the general case?
On
A space is hereditarily Lindelöf iff for every set of open sets $\mathcal{O}$ there is a countable subset $\mathcal{O'} \subseteq \mathcal{O}$ such that $\bigcup \mathcal{O}=\bigcup \mathcal{O'}$. Note that this does need any linear order on that set (which I think matters little anyway, because we can always make a union of open sets a linearly ordered one if we so wish).
Indeed a second countable space is hereditarily Lindelöf, but e.g. so is any product of separable spaces, or any ccc linearly ordered topological space, which gives us many more examples of such spaces.
No. For instance, let $X=\omega_1$ with the discrete topology, let $I=\omega_1$, and let $U_i=\{j\in\omega_1:j<i\}$. Then $\bigcup U_i=\omega_1$ but each $U_i$ is countable so the union of any countable subfamily is countable.