Existence of a numerable family given a element of a $\sigma-$algebra I'm trying to prove the next stament:

Let $E\subset\mathcal{P}(X)$ fixed. Then for all $A\in\sigma(E),$ there exists a numerable subfamily $E_{0}\subset E$ such that $A\in \sigma(E_{0}).$

My attempt is based on the follow: Let $S=\bigcup\{\sigma(E^{'}):E^{'}\subset E \quad\text{is numerable}\}.$ I've proved that $S$ is a $\sigma-$algebra. Then, I'd like to show that $E \subset S$ to prove that $S=\sigma(E)$ and finishing the proof, but I'm stuck in this.

Any kind of help is thanked in advanced.
The punchline is that ${S}$ contains ${E}$ because for any $A \in {E}$, $A \in \sigma(A)$ hence $A \in S$ by definition of $S$.
The punchline is that ${S}$ contains ${E}$ because for any $A \in {E}$, $A \in \sigma(A)$ hence $A \in S$ by definition of $S$.