# Topological space with countable open cover $\{U_\alpha\}$ with each $U_\alpha$ second-countable, is second countable

I'm trying to prove the following statement:

Let $X$ be a topological space with a countable open cover $C= \{U_\alpha : \alpha \in A\}$. If each $U_\alpha$ is second-countable then $X$ is second-countable.

However I'm stuck on how to proceed, so any hints on how to tackle it would be highly appreciated.

Let $\mathcal{B}_\alpha$ be a countable base for $U_\alpha$. Its members are by definition open in $U_\alpha$, and as all $U_\alpha$ are open in $X$, these sets are also open in $X$. So, $\mathcal{B}=\bigcup_{\alpha} \mathcal{B}_\alpha$ is a countable family of open sets in $X$. Suppose that $x \in X$ and $O$ is open in $X$ with $x \in O$. Then $x \in U_{\beta}$ for some index $\beta$. Now apply the definition of a base to see that for some $B \in \mathcal{B}_\beta$ we have $x \in B \subseteq O \cap U_\beta$. This $B \in \mathcal{B}$ and $x \in B \subseteq O$. This shows that $\mathcal{B}$ is a countable base for $X$.
Hint: Fix a countable basis $\mathcal{B}_\alpha$ for each $U_\alpha$, viewed as a subspace of $X$. There are countably many $U_\alpha$ and each of them has a countable basis. Will $\bigcup_\alpha \mathcal{B}_\alpha$ get the job done?
Is $X\subset \bigcup_{\alpha}\bigcup_{V\in \mathcal{B}_\alpha}V$? Given two elements $A,B \in\bigcup_\alpha \mathcal{B}_\alpha$ with $x\in A\cap B$ can we find $C\in \bigcup_\alpha \mathcal{B}_\alpha$ such that $x\in C\subset A\cap B$?
• @HennoBrandsma So this means after show that $\bigcup_{\alpha} \mathcal{B}_{\alpha}$ generated a topology for $X$, we have to show that this topology is equal to initial top on $X$ ?. – Sou Jan 14 '18 at 18:21