Is the countable union of second countable open subspaces second countable?

Question

I need help checking if the statement below is true. I have written a proof to it. If the statement is incorrect I'd greatly appreciate it you could point out where I went wrong in my proof. Thank you.

Let $X$ be a second countable topological space and $(X_{n})_{n\in\mathbb{N}}$ a sequences of open subspaces of $X$ with the subspace topology. If $X=\cup_{n\in\mathbb{N}}X_{n}$ and each $X_{n}$ is second countable then $X$ is second countable.

My proof:

For each $n\in\mathbb{N}$ let $B_{n}$ be a countable basis for $X_{n}$. Let $B=\cup_{n\in\mathbb{N}}B_{n}$. Notice that B is a countable union of countable sets so countable. Now, given $V$ open in $X$, for each $n \in \mathbb{N}$ we have that $V \cap X_{n}$ open in $X_{n}$. So $\exists C_{n} \subset B_{n}$ such that $V \cap X_{n} = \cup_{Y\in C_{n}}Y$. Taking $C=\cup_{n\in\mathbb{N}}C_{n}$, we have that $V = \cup_{Y\in C}Y$.

$\textbf{QED}$

Answer 1

Your collection $B$ need not be countable. Even the collectoion of all $U$ such that $U\cap X_1 \in B_1$ need not be countable.

Answer 2

Your union does not consist of open subsets of $X$, so is not a base. You'd have to assume all $X_n$ are open in $X$ for this to work.

Moreover the whole thing is false: there are countable spaces that are not second countable, so you could never proof such a "theorem".

You do have that a union of second countable subspaces has a countable network by essentially this proof. (Look at network weight here, under the topology section.)