I have proved the following statement and I would like to know if my proof is correct, thank you:
"$\mathcal{S}$, smallest $\sigma$-algebra on $\mathbb{R}$ containing $\{(r,s]:r,s\in\mathbb{Q}\}$, is the collection of Borel subsets of $\mathbb{R}$"
DEF. (Borel set): The smallest $\sigma$-algebra on $\mathbb{R}$ containing all open subsets of $\mathbb{R}$, $\mathcal{B}$, is called the collection of Borel subsets of $\mathbb{R}$. An element of this $\sigma$-algebra is called a Borel set.
LEMMA (1): Let $x\in\mathbb{R}$: then there exists both a decreasing sequence and an increasing sequence of rational numbers converging to $x$.
LEMMA (2): Every open subset of $\mathbb{R}$ can be written as a countable union of disjoint open intervals.
My proof:
Let $(a,b), a<b$ be an open interval in $\mathbb{R}$ and take $c\in\mathbb{R}, a<c<b$, arbitrary: then by Lemma (1) there exist a decreasing sequence of rational numbers $(a_i)$ such that $a_i\overset{i\to +\infty}{\to} a$ and an increasing sequence of rational numbers $(c_i)$ such that $c_i\overset{i\to +\infty}{\to} c$ so $(a_i, c_i]\in \{(r,s]:r,s\in\mathbb{Q}\}\subset\mathcal{S}$ and $(a,c]=\bigcup_{i=1}^{\infty} (a_i, c_i]\in S$ since $\mathcal{S}$, being a $\sigma$-algebra, is closed under countable unions. We thus have that every half-open interval $(a,c]\in\mathbb{R}$ belongs to $\mathcal{S}$ so it must also be that $(a,b)=\bigcup_{n=1}^{\infty}(a,b-\frac{1}{n}]\in S$ (i.e. every open interval in $\mathbb{R}$ also belongs to $\mathcal{S}$) hence, by LEMMA (2), every open set in $\mathbb{R}$. So, since every open set belongs to $\mathcal{S}$, a $\sigma$-algebra, and $\mathcal{B}$ is the smallest $\sigma$-algebra containing them by definition, we have that $\mathcal{B}\subset\mathcal{S}$. Since each half-open interval with rational endpoints $(r,s]$ belongs to $\mathcal{B}$ ($(r,s)\in\mathcal{B}$ because it is open, $\{s\}\in\mathcal{B}$ because closed sets are Borel sets so their union also belongs to $\mathcal{B}$) and $\mathcal{S}$ is the smallest $\sigma$-algebra containing them by hypothesis we also have that $\mathcal{S}\subset\mathcal{B}$ thus $\mathcal{S}=\mathcal{B}$, as desired.