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I need some help to prove that if $\mathcal{A}$ is a chain of closed subsets in a separable metric space then there is a countable subfamily $\mathcal{A}'\subseteq\mathcal{A}$ such that $\bigcup\mathcal{A}'=\bigcup\mathcal{A}$.

My attempt was to pick a countable dense subset $D$ and for each $d\in D\cap\bigcup\mathcal{A}$ choose one $A_d\in\mathcal{A}$ so that $d\in A_d$. Then the set of $A_d$'s is countable but I think that the union may not be the whole $\bigcup\mathcal{A}$. Is there any way to fix this idea or is there some better way to prove the statement?

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    Just to point something out, it is entirely possible that the dense set does not meet $\bigcup\mathcal A$. For example, let $D=\mathbb Q$, and take $A_i = \{\pi + n\mid n, finite sets are closed, and $\mathcal A=\{A_n\mid n\in\mathbb N\}$ is clearly a chain. However there is not a single rational point in $\bigcup\mathcal A$.2011-09-20

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