Let $X$ be a second countable space. I $\textbf{A}$ is a family of open set well-ordered by inclusion prove that this family is numerable.
I have the following idea: Let $\textbf{B}$ be a countable basis. Take $A_1=\min A$, we can choose $B_1\in \textbf{B}$ such that $B_1\subset A_1$ since $A_1$ is open. Then define $A_2=\min \{A-\{A_1\}\}$, we can choose $B_2\in \textbf{B}$ such that $B_2\subset A_2-\bar A_1$, since $A_2-\bar A_1$ is open. Therefore define $A_3=\min\{A-\{A_1,A_2\}\}$, the set $A_3-\bar A_2$ is open, then there exist $B_3\in \textbf{B}$ such that $B_3\subset A_3-\bar A_2$...
I want to continue this process but I don't know if it is well done. How do I formalize this idea?
Thanks!
One problem with your approach is that it could be possible that $A_2 \subseteq \overline{A_1}$, and so $A_2 \setminus \overline{A_1}$ is empty, which causes problems in picking $B_1$. It's also not exactly clear what you would do at limit stages.
Instead, note that for each $A \in \mathbf A$, if $A$ is not the maximum element, then since $\mathbf A$ is well-ordered by inclusion there is a smallest $A^+ \in \mathbf A$ such that $A \subsetneq A^+$. Now, for each non-maximum $A \in \mathbf A$ pick some $x_A \in A^+ \setminus A$, and then some $B_A \in \mathbf B$ (the chosen countable basis) such that $x_A \in B_A \subseteq A^+$. Note that for distinct $A_0 , A_1 \in \mathbf A$ it must be that $B_{A_0} \neq B_{A_1}$ (if $A_0 \subsetneq A_1$, then $A_0^+ \subseteq A_1$ and so $B_{A_0} \subseteq A_1$, however $x_{A_1} \in B_{A_1} \setminus A_1$, which gives $B_{A_0} \neq B_{A_1}$).
So we have an injection from the set of non-maximum elements of $\mathbf A$ into the countable set $\mathbf B$, meaning that there can only be countably many non-maximum elements of $\mathbf A$, and so $\mathbf A$ cannot be uncountable (our injection is undefined for at most one element of $\mathbf A$ — the maximum element, if it exists — and adding one element to a countable set cannot make the set uncountable).