We can use the notion of an analytic set to prove that not only that the Borel sets have cardinality continuum but that any countably generated $\sigma$-algebra also has cardinality continuum.

Let $Y$ be a Polish space. Recall that a subset $A\subseteq Y$ is analytic if
$A=\emptyset$ or $A=f[\mathbb{N}^\mathbb{N}]$ for some continuous $f:\mathbb{N}^\mathbb{N}\rightarrow\mathbb{R}$. We say that $A\subseteq Y$ is co-analytic if the complement $A^c$ is analytic. There are only continuumly many functions $f:\mathbb{N}^\mathbb{N}\rightarrow Y$, so there are only continuumly many analytic sets.

I claim that the collection $\mathcal{B}$ of subsets $A\subseteq Y$ that are both analytic and co-analytic form a $\sigma$-algebra that contains all closed subsets of $Y$. In fact, the set $\mathcal{B}$ is just the Borel $\sigma$-algebra, but we do not need to prove this fact to conclude that there are continuumly many Borel sets.

If $A_n\subseteq Y$ is analytic and non-empty for each $n\in\mathbb{N}$, then for all $n$, let
$f:\mathbb{N}^\mathbb{N}\rightarrow\mathbb{R}$ be a function with $f_n[\mathbb{N}^\mathbb{N}]=A_n$ for all $n$. Then
define a function $f:\mathbb{N}^\mathbb{N}\rightarrow Y$ by setting
$f((n,a_0,a_1,a_2,a_3,\dots))=f_n(a_0,a_1,a_2,\dots)$. Then
$f[\mathbb{N}^\mathbb{N}]=\bigcup_{n=0}^\infty A_n$.

To show that the collection of analytic subsets of $Y$ is closed under countable intersection, we first recall (this is not too hard to prove) that a subset $A\subseteq Y$ is analytic if and only if there is some Polish space $X$ and some continuous $f:X\rightarrow A$. Suppose now that $A_n\subseteq Y$ is analytic and non-empty for each $n$. Then for each $n$, let $X_n$ be a Polish space and let $f_n:X_n\rightarrow Y$ be a continuous function with $f_n[X_n]=A_n$. Then let
$X\subseteq\prod_{n\in\omega}X_n$ be the (possibly empty) closed subset consisting of all tuples $(x_n)_{n\in\omega}$ where $f_m(x_m)=f_n(x_n)$ for $m,n\in\omega$. Then define the continuous function $f:X\rightarrow Y$ by letting $f((x_n)_{n=0}^\infty)=f(x_0)$. Then $f[X]=\bigcap_{n=0}^\infty A_n$. Therefore, the countable intersection of analytic subsets is analytic. We may now conclude that the collection $\mathcal{B}$ of analytic and co-analytic sets is a $\sigma$-algebra. The set $\mathcal{B}$ clearly has cardinality at most continuum, so we are good.

Let $Y$ be a set and let $A_n\subseteq Y$ for all $n$. For simplicity, suppose that the sets $A_n$ are separating. Then define a function
$\iota:Y\rightarrow 2^\mathbb{N}$ by letting $\iota(y)=\{n\in\mathbb{N}:y\in A_n\}$. Then give $Y$ the coarsest topology such that the mapping $\iota$ is continuous. Then the sets $A_n$ along with the sets $A_n^c$ form a subbasis of clopen subsets of $Y$ that generate the topology on $Y$.

Let $\mathcal{C}$ be the collection of sets of the form $\iota^{-1}[A]$ for some $A\subseteq 2^\omega$ that is both analytic and co-analytic. Then $\mathcal{C}$ is a $\sigma$-algebra containing each $A_n$. Since $2^\omega$ has at most continuumly many analytic sets, the $\sigma$-algebra $\mathcal{C}$ has cardinality at most continuum. Since there are more than continuumly many Lebesgue measurable sets, if $Y=\mathbb{R}$, then $\mathcal{C}$ does not contain all the Lebesgue measurable sets.

**Going from the countable case to the continuum case**

We have established that if $\mathcal{A}$ is a countably generated $\sigma$-algebra, then $\mathcal{A}$ has cardinality continuum. We can extend this fact to show that every $\sigma$-algebra generated by continuumly many elements has cardinality continuum as well.

Suppose that $Y$ is a set, and $\mathcal{G}$ is a collection of at most continuumly many subsets of $Y$. Then for each $\mathcal{H}\subseteq P(Y)$, let
$C(\mathcal{H})$ denote the $\sigma$-algebra generated by $\mathcal{H}$.
Then $\bigcup\{C(\mathcal{H}):\mathcal{H}\subseteq\mathcal{G},|\mathcal{H}|\leq\aleph_0\}$ is a $\sigma$-algebra containing $\mathcal{G}$, but since each $C(\mathcal{H})$ has cardinality continuum and there are at most continuumly many countable subsets of $\mathcal{G}$, we conclude that $C(\mathcal{G})$ must also have cardinality continuum.

I think that most people would find the proof using the transfinite inductive construction of the $\sigma$-algebra of length $\omega_1$ to be simpler except if one wants to avoid thinking about the ordinal $\omega_1$ for some reason.