Consider an open interval $(-\infty, a)$, where $a \in \mathbb{Q}$. Why does it hold that
$$ \left\{ (-\infty, a) \mid a \in \mathbb{Q} \right\} \subset \sigma \left( \{ [x,y) \mid x,y, \in \mathbb{Q} \} \right) $$
I know that you can write $(-\infty, a)$ as $ \bigcup_{n=1}^\infty [-n, a) $ but I struggle to see how that is an element of the sigma algebra generated by some rational interval $[x, y)$. I'm assuming that you cannot just simply set $x = -n$ since $x$ is fixed.
You can simply set $x=-n$. The notation $$\{ [x,y) \mid x,y, \in \mathbb{Q} \}$$ means the set whose elements are $[x,y)$ where $x$ and $y$ range over all elements of $\mathbb{Q}$, not just some fixed elements. So the $\sigma$-algebra generated by this set contains the interval $[x,y)$ for all $x,y\in\mathbb{Q}$. In particular, for any $n\in\mathbb{N}$, we can set $x=-n$ and $y=a$ to conclude that $[-n,a)$ is in the $\sigma$-algebra.