If $E \in \sigma(\mathcal{C})$ then there exists a countable subset $\mathcal{C}_0 \subseteq \mathcal{C}$ with $E \in \sigma(\mathcal{C}_0)$
A good strategy would be to prove that $H$ is equal to $\sigma(\mathcal C)$, the $\sigma$-algebra generated by $\mathcal C$. It should be clear that $H \subset \sigma(\mathcal C)$. For the reverse inclusion, it is enough to show that $H$ is, in fact, a $\sigma$-algebra. Check each of the axioms! At some point, you will have to use the fact that a countable union of countable sets is countable.