$\coprod_{i \in I} X_i$ is second countable $\implies |I| \leq |\mathbb{N}|$

Question

Let $(X_i \neq \emptyset)_{i \in I}$ be a family topological spaces. Prove that: $\coprod_{i \in I} X_i$ is second countable $\implies |I| \leq |\mathbb{N}|$

(the coproduct is equipped with the initial topology that makes the canonical injections continuous)

My attempt:

Let $\mathcal{B}$ be a basis for the coproduct topology.

Fix $i \in I$. Then $X_i \times \{i\}$ is open in the coproduct topology.

Choose $x_i \in X_i \neq \emptyset$. Then $(x_i,i) \in X_i \times \{i\}$, and there exists $B_i \in \mathcal{B}$ such that $(x_i,i) \in B_i \subseteq X_i \times \{i\}$.

Define $\psi: I \to \mathcal{B}: i \mapsto B_i$.

Since this is an injection, it follows that $|I| \leq |\mathcal{B}|$ and the result follows.

Answer 1

The proof is correct, I would say. A nitpick: let $\mathcal{B}$ be a countable base at the start. You have actually shown that $|I| \le |\mathcal{B}|$ for any base, but the second countable assumption must be used somewhere..