Is it possible to have an uncountable strictly decreasing sequence $(A_\alpha)_{\alpha<\omega_1}$ $$A_\beta\supsetneq A_\alpha \text{ for all } \beta<\alpha$$ of closed sets $A_\alpha$ in a separable metric space $X$?
The answer is probable NO, but I need to see a proof cause I don't trust myself today!
Everyone please let me know if this sounds correct:
Suppose $(A_\alpha)_{\alpha<\omega_1}$ is a sequence of closed subsets of $X$ with $A_\beta\supseteq A_\alpha \text{ for all } \beta<\alpha$.
Then $\{X\setminus A_\alpha:\alpha<\omega_1\}$ is an open cover of $X\setminus \bigcap _{\alpha<\omega_1}A_\alpha$.
Since $X$ is hereditarily Lindelöf, the open cover has a countable subcover $\{X\setminus A_\alpha:\alpha<\gamma\}$ (covering $X\setminus \bigcap _{\alpha<\omega_1}A_\alpha$). In fact, $X\setminus \bigcap _{\alpha<\omega_1}A_\alpha\subseteq X\setminus A_\gamma$.
This means $A_\alpha=A_\beta$ for all $\gamma\leq \alpha\leq\beta$. So the sequence is not strictly decreasing.