Well-ordering on closed subsets of R I don't know how to approach the proof of the following statement:
If $A$ is a family of closed subsets of $\mathbb{R}$ well-ordered with respect to inclusion, then $A$ is countable.
Thank you in advance.
 A: The accepted answer is flawed - Maybe all rationals are in all open sets. Here's a different approach. Let $\{B_n : n < \omega\}$ list all rational intervals. Associate to each closed set $A$ the set $X_A = \{n < \omega : A \cap B_n = \phi\}$. Now check.
A: HINT: Suppose towards a contradiction that I have a family of closed sets $A_\eta$ for $\eta<\omega_1$, with $\eta<\theta$ implies $A_\eta\subsetneq A_\theta$. 


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*Show that for each successor ordinal $\eta$, there is some $x\in A_\eta$ which is not a limit of any sequence of points from $\bigcup_{\alpha<\eta}A_\alpha$. This is where we use that the $A_\eta$s are closed.

*Do you see how to use this fact to construct an uncountable set of reals which is discrete? Do you see why that's a problem?

More detail: Suppose $\{A_\theta: \theta\in\omega_1\}$ is an increasing sequence of closed subsets of $\mathbb{R}$. We'll build by induction an uncountable discrete set of reals, $D$, which can't exist. At stage $\alpha$, we define a real $x_\alpha$ and a positive real $\epsilon_\alpha$, as follows: we let $x_\alpha$ be some element of $\bigcup A_\theta$ such that $$d(x_\alpha, x_\beta)<\epsilon_\beta$$ for all $\beta<\alpha$ and $x_\alpha$ is not an accumulation point of $\{x_\beta: \beta<\alpha\}$, and $\epsilon_\alpha$ be some positive real such that the set $$\{y\in \bigcup A_\theta: d(y, x_\alpha)>\epsilon_\alpha\}$$ is uncountable.
You can show that such $x_\alpha$ and $\epsilon_\alpha$ can always be picked. But then it's clear that the set $\{x_\alpha:\alpha<\omega_1\}$ is discrete.
A: More generally, let $X$ be any hereditarily separable topological space. Assume for a contradiction that there is an uncountable family of closed subsets of $X$ which is well-ordered by inclusion. Then there is an uncountable family of closed sets of order type $\omega_1,$ i.e., there are closed sets $A_\alpha(\alpha\lt\omega_1)$ such that $A_\beta\subsetneq A_\alpha$ for $\beta\lt\alpha\lt\omega_1.$ Let $A=\bigcup_{\alpha\lt\omega_1}A_\alpha,$ and let $S$ be a countable dense subset of $A.$ Choose $\beta\lt\omega_1$ so that $S\subseteq A_\beta.$ Then $A_{\beta+1}\subseteq A\subseteq\bar S\subseteq\overline{A_\beta}=A_\beta$ contradicting $A_\beta\subsetneq A_{\beta+1}.$
