Here is an example of a $\sigma$-algebra of cardinality $2^{\aleph_0}$ which is not a subalgebra of a countably generated $\sigma$-algebra. (The previously posted examples had cardinality greater than $2^{\aleph_0}.$)
Consider the space $X=\{0,1\}^\mathbb R$ with the Tychonoff product topology, where $\{0,1\}$ has the discrete topology. Let $\mathcal A$ be the $\sigma$-algebra generated by the clopen subsets of $X.$
To see that $|\mathcal A|=2^{\aleph_0},$ we start by noting that (since $X$ is compact) a clopen set in $X$ is a finite union of basic open sets, and there are just $2^{\aleph_0}$ of those. Now the argument showing that a countably generated $\sigma$-algebra has cardinality at most $2^{\aleph_0}$ also shows that a $\sigma$-algebra generated by $2^{\aleph_0}$ sets has cardinality $2^{\aleph_0}.$
I claim that $\mathcal A$ is not contained in any countably generated $\sigma$-algebra of subsets of $X.$ Suppose $\mathcal B$ is the $\sigma$-algebra generated by a countable family $\mathcal D\subset\mathcal P(X);$ I have to show that $\mathcal A\not\subseteq\mathcal B.$ Define a mapping $f:X\to\mathcal P(D)$ by setting $f(x)=\{D\in\mathcal D:x\in D\}.$The mapping $f$ is not injective, since $|X|=2^{2^{\aleph_0}}\gt2^{\aleph_0}=|\mathcal P(D)|.$ Thus there are points $x,y\in X$ such that $x\ne y$ but $f(x)=f(y).$ Then $x$ and $y$ are not separated by any set in $\mathcal D,$ nor by any set in $\mathcal B,$ the $\sigma$-algebra generated by $\mathcal D.$ On the other hand, there is a clopen set of $X$ which separates $x$ from $y;$ that clopen set belongs to $\mathcal A$ but not to $\mathcal B.$