Show Borel sets contain all of $[a,b)$ I am struggling with how to show that a collection, $\mathcal{I}$, of integrals of the form $[a,b)$ are contained within $\mathcal{B}_\mathcal{O}$ such that $\mathcal{B}_\mathcal{O}$ is the smallest $\sigma$-algebra containing all the open sets. I have previously proved that $\mathcal{B}_\mathcal{O}$ = $\mathcal{B}_\mathcal{C}$ such that $\mathcal{B}_\mathcal{C}$ is the smallest $\sigma$-algebra containing the closed sets. I can show $\mathcal{B}_\mathcal{O}\subseteq \mathcal{B}_\mathcal{I}$. The problem is when I try to show $\mathcal{B}_\mathcal{O}\supseteq \mathcal{B}_\mathcal{I}$. Here is my thought process so far...
Let $[a_n,b_n)\in\mathcal{I}$. Then $[a_n,b_n)\in\mathcal{B}_\mathcal{I}$. Note $[a_n,b_n)=[a_n,\frac{a_n+b_n}{2}]\cup(\frac{a_n+b_n}{2},b_n)$. Notice $[a_n,\frac{a_n+b_n}{2}]\in\mathcal{C}$ and $(\frac{a_n+b_n}{2},b_n)\in\mathcal{O}$. Then $[a_n,\frac{a_n+b_n}{2}]\in\mathcal{B}_\mathcal{C}$ and $(\frac{a_n+b_n}{2},b_n)\in\mathcal{B}_\mathcal{O}$. Since $\mathcal{B}_{\mathcal{O}}=\mathcal{B}_{\mathcal{C}}$, then $[a_n,\frac{a_n+b_n}{2}]\cup(\frac{a_n+b_n}{2},b_n)\in\mathcal{B}_\mathcal{O}$ because $\mathcal{B}_\mathcal{O}$ is a $\sigma$-algebra. Therefore $\mathcal{B}_{\mathcal{O}}\supseteq\mathcal{B}_{\mathcal{I}}$.
Can I assume the union between the closed set and open set are still part of the Borel set? The closed set is countable, but is the open set countably infinite? I just want to check my thought process.

Just Note:
$$
[a,b)=\bigcap_{n=1}^{\infty}(a-\frac1{n},b)
$$
So $[a,b)$ is Borel set. Another way is
$$
[a,b)=\bigcup_{n=1}^{\infty}[a,b-\frac1{n}]
$$
Since closed set is Borel set, $[a,b)$ is Borel set.
