Countable open cover $\{U_i\}$ with each $U_i$ second countable 
Let $X$ be a topological space that admits a countable open cover $\{U_i\}_i$ such that each $U_i$ is second countable in the subspace topology. Show that $X$ is second countable. 

My attempted proof is as follows:
For each $i$, $U_i$ admits a countable basis $\mathcal{B}_i$. By the Axiom of Choice, $\cup_i \mathcal{B}_i$ is countable. All of the sets in $\cup_i\mathcal{B}_i$ are open in $X$ since they are of the form $B_i\cap U_i$ for some $B_i\in\mathcal{B}_i$, and both $B_i$ and $U_i$ are open. Since the $\{U_i\}$ are an open cover, for any $x\in X$, $x\in U_i$ for some $i$, so there is $B\in\mathcal{B}_i$ such that $x\in B$. 
Now suppose that for any $B_1,B_2\in\cup_i\mathcal{B}_i$, $x\in B_1\cap B_2$. Then there are $i,j$ such that $B_1\in\mathcal{B}_i$ and $B_2\in\mathcal{B}_j$. But I do not understand how to show there is a $B_3\in\cup_i\mathcal{B}_i$ such that $x\in B_3\subset B_1\cap B_2$. How can we show there is such a $B_3$?
I know similar questions have been asked before but I couldn't find the answer to my question.
 A: You know that the intersection $B  = B_1 \cap B_2$ is open in $X$. Since $B_1 \in \mathcal{B}_i$, we have $B \subset U_i$, i.e. $B$ is open in $U_i$. Now you find $B_3 \in \mathcal{B}_i \subset \bigcup_i \mathcal{B}_i$ such $x \in B_3 \subset U_i$.
Edit:
bof rightly remarked that you have to show something else: For each $x \in X$ and each open neigborhood $U$ of $x$ there exists $B \in \mathcal{B} = \bigcup_i \mathcal{B}_i$ such that $ x \in B \subset U$. This is done by the same argument as above: There exists $U_i$ such that $x  \in U_i$. Then $U' = U \cap U_i$ is open in $U_i$ and you find $B \in \mathcal{B}_i \subset  \mathcal{B}$ such $x \in B \subset U' \subset U$. In your question you only consider the special case $x \in U = B_1 \cap B_2$.
A: Hint: you are trying to prove something that is not necessary. Consider the collection sof all finite intersections of sets from $\cup_i \mathcal B_i$. This is  a countable collection. Show that this is a basis for the topology of $X$. 
A: There exists $U$ such that $ x \in U$. 
We know $B_1 \cap U$ and   $B_2 \cap U$ are open in $X$, and both are subsets of $U$, and also their intersection $B'$ is open. Then there exists $B_3$ a subset of $B'$ containing $x$ by the property of basis. 
