Is it true that any uncountably generated sigma algebra can be generated by a countable subset? Let $X$ be a set and let $\mathcal C$ be a non-empty collection of subsets of $X$
From my classes I was asked if the following statement is always true:

For any $A ∈ σ(\mathcal C)$, there exists a countable
sub-collection $\mathcal D ⊂ \mathcal C$ such that $A ∈ σ(\mathcal D)$.

This was not difficult to prove true, but I also considered a slight modification to the statement:

There exists a countable sub-collection $\mathcal D ⊂ \mathcal C$ such that, for any
$A ∈ σ(\mathcal C)$, $A ∈ σ(\mathcal D)$

The second statement is equivalent to saying: There exists a countable $\mathcal D ⊂ \mathcal C$ such that $σ(\mathcal C) = σ(\mathcal D)$. My intuition is that this is false but I can't think of any examples at least in the case for $X=\mathbb R$. So my question is this, is the modified statement true for $X=\mathbb R$ and if yes then is it also true for arbitrary $X$.
 A: No, the statement is not even true for $X=[0,1)$. Let $\mathcal{C}= \{\{x\}|\;x\in [0,1]\setminus \mathbb{Q}\}$. We'll prove that $\sigma(\mathcal{C})$ is not countably generated. One may note that every set in this $\sigma$-algebra must have one of the forms
\begin{align}
&\emptyset\\
&\ [0,1)\\
&\bigcup_{n=1}^{\infty}\{x_n\}\\
&\bigcap_{j=1}^{\infty} [0,1)\setminus \{y_j\}\\
&(\bigcup_{n=1}^{\infty}\{x_n\})\cup(\bigcap_{j=1}^{\infty} [0,1)\setminus \{y_j\})
\end{align}
for any choice of sequences of irrationals $x_n$ and $y_j$, since taking unions and complements of the last type of set yields a set of one of the previous forms.
For $x\in [0,1)$, define the atom of x, $A(x):=\cap_{x\in E\in \sigma(\mathcal{C})} E$. If $\sigma(\mathcal{C})$ were countably generated, we would have $A(x)\in \sigma(\mathcal{C})$ for every $x$. To see this, note that if $\sigma(\mathcal{C})=\sigma(\mathcal{D})$, where $\mathcal{D}$ is countable, we would have
$$
A(x)=(\bigcap_{x\in D\in \mathcal{D}} D)\cap (\bigcap_{x\not\in D\in \mathcal{D}} [0,1)\setminus D):=A'(x)\in \sigma(\mathcal{D})
$$
To see this, note that $\mathcal{E}=\{E\subseteq [0,1)|\; x\in E\Longrightarrow A'(x)\subseteq E\}$ is a $\sigma$-algebra containing $\mathcal{D}$.
Now, clearly, for $x\in [0,1)\setminus \mathbb{Q}$, we have $A(x)=\{x\}\in \mathcal{C}$, so assume $x\in \mathbb{Q} \cap  [0,1)$. Using the above characterisation of $\sigma(\mathcal{C})$, we easily get that $A(x)=\mathbb{Q}\cap [0,1)$ and that $\mathbb{Q}\cap [0,1)$ is not an element of $\sigma(\mathcal{C})$. Thus, $\sigma(\mathcal{C})$ cannot be countably generated even though it's a sub-$\sigma$-algebra of the Borel algebra.
