= Solution
Let $A=\bigcup_{n\ge1}A_n$, and suppose $A=P\cup Q$ were a separation into disjoint nonempty relatively open sets. Each <connected subset> $A_n$ lies wholly in $P$ or wholly in $Q$, by the argument just proved. Since $A_n\cap A_{n+1}\ne\varnothing$, consecutive subsets must lie on the same side. Induction puts every $A_n$ on the side containing $A_1$. The other side is then empty, a contradiction. Hence \b[$A$ is connected]. This is the <connected chain of overlapping subsets> argument; it does not require all the subsets to share one common point.
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