Q:

Conclude that $\sigma(\mathcal{B}_0) = \mathcal{B}(\mathbb{R})$

Suppose that $B_0 = \{(-\infty, x): x\in\mathbb{R}\}$ and let $\mathcal{B}(\mathbb{R})$ be the Borel $\sigma$-algebra of $\mathbb{R}$.
Previously I've shown that:

$\sigma(B_0)$ contains all open intervals.
Every open set in $\mathbb{R}$ can be written as the union of countably many open intervals.

Now I want to conclude, based on what I showed previously, that $\sigma(\mathcal{B}_0) = \mathcal{B}(\mathbb{R})$.
What I've tried: I think I need to show that $\mathcal{B}(\mathbb{R}) \subseteq \sigma(B_0)$ and $\sigma(B_0) \subseteq \mathcal{B}(\mathbb{R})$. 
I think I've found a way to show that $\mathcal{B}(\mathbb{R})\subseteq \sigma(B_0)$: every open set in $\mathbb{R}$ can be written as the union of countable many open intervals. Since $\sigma(B_0)$ contains all open intervals we know that for every set $A\in\mathcal{B}(\mathbb{R})$, there exist $A_1,A_2, ..., \in \sigma(B_0)$ such that $\bigcup\limits_{j =1}^{\infty}A_j  = A$, hence $A\in \sigma(B_0)$. 
Question: How do I show that $\sigma(B_0) \subseteq \mathcal{B}(\mathbb{R})$?

A:

Since each member of $B_0$ is open, we have $B_0 \subseteq \mathcal{B}(\mathbb{R})$. 
Hence
$$\sigma(B_0) \subseteq \sigma(\mathcal{B}(\mathbb{R})) = \mathcal{B}(\mathbb{R}).$$

