I am trying to prove that the cardinality of the Borel $\sigma$-algebra on $\mathbb{R}$ equipped with the standard topology is $c$, where $c=|\mathbb{R}|$ is the cardinality of the continuum.
My attempt
The Borel $\sigma$-algebra is the smallest $\sigma$-algebra containing the open sets of $\mathbb{R}$ in the standard topology. Since every open set of $\mathbb{R}$ is the countable disjoint union of open intervals, it follows that $\mathcal{B}=\sigma(\mathcal{C})$ where $\sigma(\mathcal{C})$ is the $\sigma$-algebra generated by $\mathcal{C}$ where $\mathcal{C}$ is the set of all open intervals of $\mathbb{R}$.
It is clear to me that we should proceed to prove the result by constructing the Borel $\sigma$-algebra, but it is here where I run into some problems/misunderstandings:
Let us start with the set of all open intervals $(a,b)$. Denote by $B_0$ this collection. Now, for a successor ordinal $\alpha$ (with predecessor $\alpha-1$), let $B_{\alpha}$ be the closure of $B_{\alpha-1}$ with regards to complementation, and countable union (including the subsets obtained via complementation). For the case where $\alpha$ is a limit ordinal, let $B_{\alpha}$ be the union of all previous sets constructed in this manner. Terminate at the first uncountable ordinal ($\omega_1$).
We want to show this is a $\sigma$-algebra. So, we need to show that the constructed $\mathcal{B}$ which is the union of all the collections $B_i$ we constructed in the above manner contains $\emptyset, \mathbb{R}$, and is closed under complementation/countable unions.
Clearly $\mathbb{R}\in\mathcal{B}$ since $\mathbb{R}=\bigcup\limits_{q\in\mathbb{Q}}(q,q+1)$. Also, $\emptyset\in\mathcal{B}$ since it is the complement of $\mathbb{R}$ (in particular, this is because it would have been added in the second iteration of the induction as per the process outlined above).
Next we show closure under complementation. Let $A\in\mathcal{B}$. then $A\in B_{\alpha}$ for some ordinal $\alpha$. Since every ordinal has an immediate successor, we have that $A^c\in B_{\alpha+1}\subseteq\mathcal{B}$ as desired. Showing closure under countable unions is identical. It remains to show that the generated collection is precisely the Borel $\sigma$-algebra, $\text{Bor}$. Since every open set is added in the second induction step (every open set is a disjoint countable union of open intervals), we have clearly that $\text{Bor}\subseteq\mathcal{B}$. As well, every open interval is in $\text{Bor}$, so under closure, $\text{Bor}$ must contain all of $\mathcal{B}$.
Now, the cardinality of the collection of all open intervals is $c$. It remains to show that at each stage in the induction we never add more than $c$ things, and the result will follow. In the complementation step we clearly only add $c$ more things, and $c+c=c$. In the union step we add only the countable Cartesian product of sets having cardinality $c$ many element, but it is well known that this has cardinality at most $c$ again. So clearly at every induction step we add no more than $c$ many elements, which gives us that $|B_{\alpha}|=c$ for every ordinal $\alpha<\omega_1$.
To construct $\mathcal{B}$ we take the union of these sets each with cardinality $c$. But the $c$-union of cardinality $c$ sets again has cardinality $c$ and the proof is complete.
Does this proof sound about right? Do you think I need any more details? Is there maybe a nicer way to word certain things (I feel as if I got a bit hand-wavy towards the end)?