As you say, $\mathcal{A}$ is the set of subsets $A\subseteq \Omega$ such that $A$ or $A^c$ is at most countable. The measure $P:\mathcal{A}\rightarrow [0,1]$ is: $P(A)=0$ if $A$ is at most countable and $P(A)=1$ if $A^c$ is at most countable.
If $X:\Omega\rightarrow\mathbb{R}$ is measurable then I claim that there exists $r\in\mathbb{R}$ such that $X^{-1}(r)$ is uncountable. In fact, suppose that $X^{-1}(r)$ is at most countable for all $r$. Then $X(\Omega)$ is uncountable and therefore there exists $a\in \mathbb{R}$ such that $X(\Omega)\cap (-\infty,a)$ and $X(\Omega)\cap (a,\infty)$ are uncountable. If $A=X^{-1}((-\infty,a))$ then $A^c=X^{-1}((a,\infty))$ and it is clear that both $A$ and $A^c$ are uncountable, which is a contradiction.
So, let $X:\Omega\rightarrow\mathbb{R}$ measurable and let $r\in\mathbb{R}$ such that $A=X^{-1}(r)$ is uncountable. Then $P(A^c)=0$ and therefore
$$E[X,\mathcal{A}]=\int_{\Omega} XdP=\int_AXdP=rP(A)=r$$
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 $E^c \cap \Omega \in A_1$.
$$ $$
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 $n$, $\cup_{n \in \mathbb{N}} (E_n \cap \Omega) \in A_1$.
$$ $$
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 $x \in \Omega \wedge x \notin E$.
$$ $$
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
To show $\mathcal{B} \subseteq \sigma(\mathcal{C})$, let $E \in \mathcal{B}$. Then either $E$ is countable or $E^{c}$ is countable.
But if $E$ is countable, then $E = \{ x_{1}, x_{2}, \dots \}$ where $x_{i} \in \Omega$. But this means $E = \bigcup \limits_{i = 1}^{\infty} \{ x_{i} \}$, and since $\{ x_{i} \}$ is in $\sigma( \mathcal{C})$ for each $i$ (since $\{ x_{i} \} \in \mathcal{C}$ for each $i$), then the countable union $\bigcup \limits_{i = 1}^{\infty} \{ x_{i} \}$ is in $\sigma(\mathcal{C})$ (since $\sigma(\mathcal{C})$ is a $\sigma$-algebra and thus closed under countable unions), and so $E \in \sigma(\mathcal{C})$.
If, however, $E^{c}$ is countable, then by the same argument above, we get that $E^{c} \in \sigma(\mathcal{C})$, and since $\sigma(\mathcal{C})$ is a $\sigma$-algebra (and thus closed under complements), this gives that $E \in \sigma(\mathcal{C})$, as desired.
So, we've shown that given $E \in \mathcal{C}$, if it is countable, it is in $\sigma(\mathcal{C})$, and if its complement is countable, $E$ is still in $\sigma(\mathcal{C})$, which implies $\mathcal{B} \subseteq \sigma(\mathcal{C})$.