Disjoint Union of First Countable Spaces This question concerns part $d)$ of proposition 3.42 of Lee's book on topological manifolds.

Let $ \left( X_{\alpha} \right)_{\alpha \in A} $ be an (arbitrary) indexed family of topological spaces. If each $X_{\alpha}$ is first countable, then $X = \coprod_{\alpha \in A} X_{\alpha}$ is first countable.


My attempt:
For each $\alpha$, let $\mathcal{B}_{p_{\alpha}}$ be a neighborhood basis for $p_{\alpha} \in X_{\alpha}$ and take $\mathcal{B}_{p} = \coprod_{\alpha \in A} \mathcal{B}_{p_{\alpha}}$. It's not hard to show that for any $U \subseteq X$ neighborhood of $p$, there is a $B \in \mathcal{B}_p$ with $B \subseteq U$.
Consider now the bijection $\psi_{\alpha} : \mathcal{B}_{p_{\alpha}} \to \mathbb{N}$ and the canonical injection $\iota_{\alpha} : \mathcal{B}_{p_{\alpha}} \to \mathcal{B}_p$ given by $B_i \mapsto \left(B_i, \alpha\right)$. Take $g = \iota_{\alpha} \circ \psi^{-1}_{\alpha} : \mathbb{N} \to \mathcal{B}_p$. It is clear that $g$ is injective.
The goal now would be to construct or show the existence of an injection from $\mathcal{B}_p $ to $\mathbb{N}$ and then proceed to use Schroeder-Bernstein's theorem. If there is an injection $\phi_{\alpha}: \mathcal{B}_p \to \mathcal{B}_{p_{\alpha}}$, then $f = \psi_{\alpha} \circ \phi_{\alpha}: \mathcal{B}_p \to \mathbb{N}$ would be the desired injection. However, the map $\iota^{-1}_{\alpha}$ is not injective. Is there such an injection? Is this approach salvageable? If not, which approach would work?
 A: There is a problem in that you haven't defined $\mathcal{B}_p$. I am also concerned that you think the disjoint union space is like the product space, in that $p\in X$ is really equal to a family $(p_\alpha\in X_\alpha)_{\alpha\in A}$. This is not so.
You don't even really need to worry about setting up injections, bijections (but you can if you want to).

It's not hard to show that for any $U\subseteq X$ a
neighbourhood of $p$, there is a $B\in\mathcal{B}_{p_\alpha}$ with $B\subseteq U$.

With this remark alone, you are essentially done. Any $p\in X$ is necessarily a $p\in X_{\alpha}$ for some $\alpha$. Fixing this $\alpha$, there is a countable local base $\mathcal{B}_{p_\alpha}$ for the topology of $X_\alpha$ at $p$. You recognise that every element of this basis remains open in $X$, and you also recognise that any neighbourhood of $p\in X$ contains an element of this basis. By very definition, $\mathcal{B}_{p_\alpha}$ is a countable basis for the topology of $X$ at $p$. As $p$ is arbitrary, you are done.
