Show that the collection of Borel sets in $\mathbb{R}$ has the same cardinality $c$ as $\mathbb{R}$ does.
Hint: Show easily that there are at least $c$ Borel sets. Then take an uncountable well-ordered set $(J,<)$ such that for each $j\in J,\{i\in J:i<j\}$, is countable. Let $\alpha$ be the least element of $J$. Recursively, define families $\mathcal{B}_j$ of Borel sets in $\mathbb{R}$: let $\mathcal{B}_\alpha$ be the collection of all open intervals with rational endpoints. For any $\beta\in J$, if $\beta'$ is the next larger element, recursively let $\mathcal{B}_{\beta'}$ be the collection of all complements and countable unions of sets in $\mathcal{B}_\beta$. If $\gamma\in J$ does not have an immediate predecessor ($\gamma\neq\beta'$ for all $\beta$), $\gamma>\alpha$, let $\mathcal{B}_\gamma$ be the union of $\mathcal{B}_\beta$ for all $\beta<\gamma$. Show that the union of all the $\mathcal{B}_\beta$ for $\beta\in J$ is the collection of all Borel sets, and so that its cardinality is $c$ using the following lemma.
Lemma 0 Let $X$ be a non-empty set of cardinality less than or equal to $c$ and let $Y$ have cardinality $c$. Then $X\times Y$ has cardinality $c$.
My efforts:
$\{x\}$ is closed and thus Borel for any $x\in\mathbb{R}$. Therefore, there are at least $c$ Borel sets.
Then take an uncountable well-ordered set $(J,<)$ such that for each $j\in J,\{i\in J:i<j\}$, is countable. Let $\alpha$ be the least element of $J$. Recursively, define families $\mathcal{B}_j$ of Borel sets in $\mathbb{R}$: let $\mathcal{B}_\alpha$ be the collection of all open intervals with rational endpoints. For any $\beta\in J$, if $\beta'$ is the next larger element, recursively let $\mathcal{B}_{\beta'}$ be the collection of all complements and countable unions of sets in $\mathcal{B}_\beta$. If $\gamma\in J$ does not have an immediate predecessor ($\gamma\neq\beta'$ for all $\beta$), $\gamma>\alpha$, let $\mathcal{B}_\gamma$ be the union of $\mathcal{B}_\beta$ for all $\beta<\gamma$. Show that the union of all the $\mathcal{B}_\beta$ for $\beta\in J$ is the collection of all Borel sets, and so that its cardinality is $c$.
Let $\mathcal{B}$ be the collection of Borel sets in $\mathbb{R}$. We will show $\bigcup_{\beta\in J}\mathcal{B}_\beta=\mathcal{B}$.
We first show $\bigcup_{\beta\in J}\mathcal{B}_\beta\subset\mathcal{B}$. We will show $\mathcal{B}_\beta\subset\mathcal{B}$ for each $\beta\in J$ by transfinite induction. The base case obviously holds: $\mathcal{B}_\alpha\subset\mathcal{B}$. Next we need to show the successor case. Assume $\mathcal{B}_\beta\subset\mathcal{B}$. Then each set in $\mathcal{B}_\beta$ is Borel. Let $\beta'$ be the immediate successor of $\beta$. Let $B'\in\mathcal{B}_{\beta'}$. Then $B'$ is either complement of a set in $\mathcal{B}_\beta$ or countable union of sets in $\mathcal{B}_\beta$. In either case, $B'$ is Borel since complement of Borel set is Borel and countable union of Borel sets is Borel. Finally we need to show the limit case. Assume $\gamma\in J$ does not have an immediate predecessor ($\gamma\neq\beta'$ for all $\beta$), $\gamma>\alpha$. Also assume $\mathcal{B}_\beta\subset\mathcal{B}$ for all $\beta<\gamma$. Then $\mathcal{B}_\gamma=\bigcup_{\beta<\gamma}\mathcal{B}_\beta\subset\mathcal{B}$.
Next we will show that $\mathcal{B}\subset\bigcup_{\beta\in J}\mathcal{B}_\beta$. We need the following theorem.
Theorem 1 $\mathcal{B}$ is the smallest $\sigma$-algebra containing $\mathcal{B}_\alpha$.
Obviously $\mathcal{B}_\alpha\subset\bigcup_{\beta\in J}\mathcal{B}_\beta$. By Theorem 1, we only need to show $\bigcup_{\beta\in J}\mathcal{B}_\beta$ is a $\sigma$-algebra. First we need to show $\mathbb{R}\in\bigcup_{\beta\in J}\mathcal{B}_\beta$. This follows from the fact that $(-n,n)\in\mathcal{B}_\alpha$ for each positive integer $n$ and the fact that $\mathbb{R}=\bigcup_{n=1}^\infty(-n,n)\in\mathcal{B}_{\alpha'}$ where $\alpha'$ is the immediate successor of $\alpha$. Next we need to show $\bigcup_{\beta\in J}\mathcal{B}_\beta$ is closed under complements. Let $B\in\bigcup_{\beta\in J}\mathcal{B}_\beta$. Then there exists a $\beta\in J$ such that $B\in\mathcal{B}_\beta$. Let the complement of $B$ be $B^C$. By definition, $B^C\in\mathcal{B}_{\beta'}\subset\bigcup_{\beta\in J}\mathcal{B}_\beta$ where $\beta'$ is the immediate successor of $\beta$. Finally we need to show $\bigcup_{\beta\in J}\mathcal{B}_\beta$ is closed under countable unions. Let $B_1,B_2,\cdots\in\bigcup_{\beta\in J}\mathcal{B}_\beta$ and $B=\bigcup_{n=1}^\infty B_n$. Then there exists $\beta_i\in J$ such that $B_i\in\mathcal{B}_{\beta_i}$ for each $i$. We need to show $B\in\bigcup_{\beta\in J}\mathcal{B}_\beta$. We need the following theorem.
Theorem 2 If $A$ is a countable subset of $J$, then $A$ has an upper bound in $J$.
By Theorem 2, $\{\beta_1,\beta_2,\cdots\}$ has an upper bound in $J$, say $\gamma$. We will prove the following lemma.
Lemma 1 $\mathcal{B}_\theta\subset\mathcal{B}_\phi$ if $\theta<\phi$.
Proof We will show this in two cases: (1) There exists a chain of immediate successors $\theta<\cdots<\phi$; (2) There does not exist a chain of immediate successors $\theta<\cdots<\phi$. In case (1), the lemma follows by definition: $\mathcal{B}_\beta\subset\mathcal{B}_{\beta'}$ where $\beta'$ is the immediate successor of $\beta$. In case (2), $\phi\in J$ does not have an immediate predecessor ($\gamma\neq\beta'$ for all $\beta$), $\gamma>\alpha$. By definition, $\mathcal{B}_\theta\subset\bigcup_{\beta<\phi}\mathcal{B}_\beta=\mathcal{B}_\phi$.
By Lemma 1, $\mathcal{B}_{\beta_i}\subset\mathcal{B}_\gamma$ for each $i$. Therefore, $B_i\in\mathcal{B}_\gamma$ for each $i$. Thus $\mathcal{B}=\bigcup_{n=1}^\infty B_n\in\mathcal{B}_\gamma\subset\bigcup_{\beta\in J}\mathcal{B}_\beta$.
We have shown $\bigcup_{\beta\in J}\mathcal{B}_\beta=\mathcal{B}$. Finally we need to show the cardinality of $\bigcup_{\beta\in J}\mathcal{B}_\beta$ is $c$. Obviously $\mathcal{B}_\alpha$ is countable. Next we will show that $\mathcal{B}_{\beta'}$ is countable if $\mathcal{B}_\beta$ is countable where $\beta'$ is the immediate successor of $\beta$. Let $\mathcal{C}_\beta$ be the collection of all complements of sets in $\mathcal{B}_\beta$. Obviously $\mathcal{C}_\beta$ is countable. Let $\mathcal{B}_\beta(n)$ be the collection of unions of $n$ sets in $\mathcal{B}_\beta$. $\mathcal{B}_\beta(n)$ is countable for each $n$ since the set of all $n$-tuples of integers is countable. Then $\mathcal{B}_{\beta'}=\mathcal{C}_\beta\cup\bigcup_{n=1}^\infty\mathcal{B}_\beta(n)$ is countable since countable union of countable set is union. $\mathcal{B}_\gamma$ for $\gamma$ without immediate predecessor is also countable by definition since $\{\beta\in J:\beta<\gamma\}$ is countable.
My questions:
(1) The recursive definition should define $\mathcal{B}_\beta$ all $\beta$. Why do we need additional definition for $\gamma$ without immediate predecessor?
(2) How to show the cardinality of $\bigcup_{\beta\in J}\mathcal{B}_\beta$ is $c$ using Lemma 0?
It's easier on notation just to choose the canonical $\omega_1$ ordinal as $J$. Then $\mathcal{B}_0$ is just the rational intervals as the starting level (and is even countable so less than size continuum, $\mathfrak{c}$).
If $\mathcal{B}_\alpha$ has size at most $\mathfrak{c}=2^{\aleph_0}$ then there are at most $\mathfrak{c}=(2^{\aleph_0})^{\aleph_0}$ many sequences of subsets from $\mathcal{B}_\alpha$ to take a union of, so the set $\mathcal{B}_{\alpha+1}$ consisting of the complements of the sets from $\mathcal{B}_\alpha$ and the countable unions of sequences from $\mathcal{B}_\alpha$ also has size at most $\mathfrak{c} + \mathfrak{c} = \mathfrak{c}$.
And if $\gamma < \omega_1$ is a limit ordinal and all $\mathcal{B}_\alpha$ where all $\alpha < \gamma$ have size at most $\mathfrak{c}$, their union $\mathcal{B}_\gamma$ has size at most $|\gamma|\cdot \mathfrak{c} = \mathfrak{c}$ as well.
So this shows by induction that $|\mathcal{B}_\alpha| \le \mathfrak{c}$ for all $\alpha < \omega_1$ and so $\mathcal{B}_{\omega_1} = \bigcup_{\alpha < \omega_1} \mathcal{B}_\alpha$ has size at most $\aleph_1 \cdot \mathfrak{c} = \mathfrak{c}$ as well.
Now, $\mathcal{B}_{\omega_1}$ is closed under complements and countable unions (if $F \in \mathcal{B}_{\omega_1}$ we know that for some $\alpha < \omega_1$, $F \in \mathcal{B}_{\alpha}$ and then $F^\complement \in \mathcal{B}_{\alpha + 1 } \subseteq \mathcal{B}_{\omega_1}$ by definition. Similarly for countable unions, using that a sequence $\alpha_n < \omega_1$ has an upper bound $\beta < \omega_1$ and the $\mathcal{B}_\alpha$ are increasing (your lemma 1).
And indeed $\Bbb R \in \mathcal{B}_1$, as a countable union of open intervals $(-n,n)$ say, so $\mathcal{B}_{\omega_1}$ is a $\sigma$-algebra and all open sets are in it, as they are already in $\mathcal{B}_{1}$ (all open sets are countable unions of rational intervals), so by minimality $\textrm{Bor}(\Bbb R) \subseteq \mathcal{B}_{\omega_1}$ and the reverse is true by an easy induction, that you already did. It follows that there are exactly $\mathfrak{c}$ many Borel sets.