A space with a sequence of metrizable subspaces is metrizable.

Question

A compact Hausdorff space $X$ has a sequence of subspaces $A_1\subset A_2 \subset A_3 \subset \cdots $.
And $A_n$ has the following properties
・$\cup A_n =X$
・All $A_n$ are metrizable (as a relative topology of $X$).
In this case, prove that $X$ is metrizable

I have almost completed the proof of this problem. However, there are two gaps. If you know the proof, please let me know.

Here is my proof

1. From Urysohn's lemma, it is equivalent that a compact Hausdorff space is metrizable and that it is a second-countable space

2. Let $d_i$ be the distance of $A_i$ The distance $d'_i$ in $\overline{A_i}$ is defined as follows.
   When $x,y$ are in $A_i$, $d'(x,y) = d(x,y)$
   When $x$ is in $A_i$ and $y$ is not in $A_i$, there exists $A_j$ such that $y$ is in $A_j$. In this case, for any natural number $n$, there exists $y_n$ such that $d(y, y_n) ≤ 1/n$, so $d'(x, y) = \lim_{n→∞} d(x, y_n)$
   When $x,y$ do not belong to $A_i$, we construct the sets $d'(x,y) = \lim_{n→∞} d(x_n,y_n)$
   This is well-defined and satisfies the distance condition (we have not yet proved this).

3. Therefore, $\overline{A_i}$ is metrizable. Since it is compact, it is also a second-countable space.

4. From the countable basis of $\overline{A_i}$, we can construct a countable basis for $X$ (this has not yet been proven)

5. Therefore, since $X$ is a second-countable space, it is metrizable.

The following two points have not yet been completed in this proof
・The proof that the definition of distance in $\overline{A_i}$ is well-defined and satisfies the conditions for distance
・The method for constructing a countable basis for X from the countable basis for $\overline{A_i}$

If you know how to do either of these, please let me know. Also, if you know of any other approaches to this problem, please let me know.

Answer 1

This follows from known theorems about metrizability of union of metrizable subspaces.

Def. $M\subseteq X$ is locally dense when there is open $U\subseteq X$ with $M\subseteq U$ such that $M$ is dense in $U$. In particular if $M$ is dense, then it's locally dense.

Theorem 1.1 (Metrizability of certain countable unions by Corson and Michael) If $X$ is a $T_4$ space such that $X = \bigcup \mathfrak{M}$ where each $M\in \mathfrak{M}$ is locally dense and metrizable, and $\mathfrak{M}$ is countable, and if one of the following is true:

• $X$ is separable
• $X$ is locally compact
• $X$ is $\sigma$-compact
• $X$ is a $G_\delta$-space
• For each $M\in\mathfrak{M}$ there is $F_\sigma$-set $A_M\subseteq M$ of $X$ with $\bigcup\{A_M : M\in\mathfrak{M}\} = X$
• $\mathfrak{M}$ is finite and for each $M\in\mathfrak{M}$ there is $G_\delta$-set $A_M\subseteq M$ of $X$ with $\bigcup\{A_M : M\in\mathfrak{M}\} = X$
• $\mathfrak{M}$ is finite and for each $M\in\mathfrak{M}$ there is completely metrizable $A_M\subseteq M$ of $X$ with $\bigcup\{A_M : M\in\mathfrak{M}\} = X$
• $\mathfrak{M}$ has two elements and $X$ is absolute $G_\delta$

then $X$ is metrizable.

Theorem VI.13 (Modern general topology by Nagata) If $X$ is a collectionwise normal, locally countably compact space such that $X = \bigcup\mathfrak{M}$ where each $M\in\mathfrak{M}$ is closed and metrizable, and $\mathfrak{M}$ is countable, then $X$ is metrizable.

Corollary If $X$ is compact Hausdorff and $X = \bigcup_n A_n$ where each $A_n$ is metrizable and $A_n\subseteq A_m$ for $n\leq m$, then $X$ is metrizable.

Proof: Since $\overline{A_n}\cap A_m\supseteq A_n$ for $m\geq n$ are dense metrizable subspaces of $\overline{A_n}$ such that $\bigcup_{m\geq n} (\overline{A_n}\cap A_m) = \overline{A_n}$, it follows from theorem 1.1 that $\overline{A_n}$ is metrizable. From theorem VI.13 it follows that $X$ is metrizable. $\square$

Remarks.

1. The condition that $A_n\subseteq A_m$ for $n\leq m$ is essential, since if $X$ is the one-point compactification of uncountable discrete space, then $X$ is a union of two metrizable subspaces, but not metrizable.

2. The condition that $X$ is Hausdorff is essential, since if $X$ is any compact countable $T_1$ space which is not Hausdorff, then writing $X = \{q_n : n\in\mathbb{N}\}$ we can let $A_n = \{q_1, ..., q_n\}$.

3. The condition that $X$ is compact is essential, since if $X$ is any countable Hausdorff non-metrizable space, and $X = \{q_n : n\in\mathbb{N}\}$ then we can take $A_n = \{q_1, ..., q_n\}$.

4. The condition that $X$ is compact Hausdorff can be replaced by locally compact collectionwise $T_4$. Any locally compact Hausdorff paracompact space is collectionwise $T_4$. It's a set theory issue whetever or not every locally compact $T_4$ space is collectionwise $T_4$.

5. The condition that $X$ is compact Hausdorff cannot be replaced by locally compact Hausdorff, because if $X = \Psi(\mathbb{N}, \mathcal{A})$ is Isbell-Mrówka space where $\mathcal{A}$ is a maximal almost disjoint family on $\mathbb{N}$, then $X$ is locally compact and non-metrizable, but $A_n = \{1, ..., n\}\cup \mathcal{A}$ are closed metrizable subspaces of $X$ such that $\bigcup_n A_n = X$.