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 $A_1$.
As you said "All open sets $\mathcal{O}$ are complements of closed sets. Since $\sigma(\mathcal{C})$ contains complements of closed sets, then $\mathcal{O}\subset\sigma(\mathcal{C})$".
By definition, $\sigma(\mathcal{O})$ is the smallest $\sigma$-algebra containing $\mathcal{O}$. As $\sigma(\mathcal{C})$ is a $\sigma$-algebra it follows that $\sigma(\mathcal{O})\subseteq \sigma(\mathcal{C})$. Similarily, $\mathcal{C}\subseteq \sigma(\mathcal{O})$ and therefore by the same argument, $\sigma(\mathcal{C})\subseteq \sigma(\mathcal{O})$. It follows that $\sigma(\mathcal{C})=\sigma(\mathcal{O})$.
Best Answer
Consider $\{X \in B: X \cap G \in A_1\}$. Verify that this is sigma algebra. It contains all open sets in $\mathbb R$. Hence it contains all Borel sets. In particular, if $X$ is Borel and contained in $G$ then $X \in A_1$.