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.
To prove that $A_2 \subset A_1$, first, recall that $\mathbb{B} = \sigma(\text{open subsets of }\mathbb{R}).$ We already know that the intersection of any open subset of $\mathbb{R}$ with $\Omega$ is contained in $A_1$. From here, to prove that the intersection of any Borel set of $\mathbb{R}$ is contained in $A_1$, it suffices to prove the following two claims:
(1) If $\{E_n \cap \Omega\}_{n \in \mathbb{N}} \subset A_1$, then $$ \left(\cup_{n \in \mathbb{N}}E_n \right) \cap \Omega \in A_1 \text{,}$$
i.e. if $\{E_n\}$ is an indexed collection of sets so that the intersection of each with $\Omega$ is in $A_1$, then the intersection of their union with $\Omega$ is contained in $A_1$.
(2) If $E \cap \Omega \in A_1$, then $E^c \cap \Omega \in A_1$.
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Proof of (1):
Let $\{E_n\}$ be as in the statement. Then
$$\left( \cup_{n \in \mathbb{N}} E_n \right) \cap \Omega = \cup_{n \in \mathbb{N}} (E_n \cap \Omega).$$ Since $A_1$ is a $\sigma$-algebra on $\Omega$ and $E_n \cap \Omega \in A_1$ for all $n$, $\cup_{n \in \mathbb{N}} (E_n \cap \Omega) \in A_1$.
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Proof of (2):
Let $E$ be a subset of $\mathbb{R}$ so that $E \cap \Omega \in A_1$. Since $A_1$ is a $\sigma$-algebra on $\Omega$ (and hence closed under complementation relative to $\Omega$), we have that $\Omega \backslash(E \cap \Omega) \in A_1$.
We claim that $E^c \cap \Omega = \Omega \backslash(E \cap \Omega)$. The inclusion $E^c \cap \Omega \subset \Omega \backslash(E \cap \Omega)$ is obvious. To prove the reverse inclusion, suppose that $x \in \Omega \backslash(E \cap \Omega)$. Then $x \in \Omega \wedge (x \notin (E \cap \Omega))$, i.e. $x \in \Omega \wedge (x \notin E \lor x \notin \Omega)$. The only way for this statement to be true is to have $x \in \Omega \wedge x \notin E$.
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To apply these claims, let $\mathcal{C}$ be the collection of sets $E \subset \mathbb{R}$ so that $E \cap \Omega$ is in $A_1$. The claims (1) and (2), combined with the fact that $A_1$ contains any set that is open relative to $\Omega$, show that $\mathcal{C}$ is a $\sigma$-algebra that contains the open sets of $\mathbb{R}$. Hence, by definition of $\mathbb{B}$, $\mathbb{B} \subset \mathcal{C}$. As a result, the intersection of any Borel set with $\Omega$ is contained in $A_1$.
Best Answer
First of all, $\{\varnothing,\Omega,C_1,\lnot C_1\}$ is not a $\sigma$-algebra on $\Omega$. Recall that $C_1$ is a collection of subsets of $\Omega$ and a $\sigma$-algebra on $\Omega$ is a collection of subsets of $\Omega$. So $C_1$ should be a subset of $\sigma(C_1)$ and not an element of it.
You can define $\sigma(C)$ as the intersection of all $\sigma$-algebras which contain $C$. But you can also start closing $C$ under complements and countable unions. The problem is that this process might take you $\omega_1$ steps to complete (and without the axiom of choice, possibly longer).
So while it is very very useful to understand the internal construction of "the smallest $\sigma$-algebra such that ...", it is unnecessary in this case. The second definition is easier to use.
HINT: Note that if $C_2\subseteq\sigma(C_1)$ then $\sigma(C_2)\subseteq\sigma(C_1)$ (by definition of "smallest").