A simple function is of the form $$f(x) = \sum_{i = 1}^N a_i 1_{A_i}$$ a finite linear combination of functions $1_{A_i}$ for $A_i \in \sigma(A)$.
We would like to prove that those functions are in $\mathcal{H}$.
So we consider $\mathcal{C} = \{B \in \sigma(A)\mid 1_B \in \mathcal{H}\}$
So $\mathcal{A} \subset \mathcal{C}$ by $(i)$.
Since $\Omega \in \mathcal{A}$, $\Omega \in \mathcal{C}$
Moreover, if $A \subset B \in \mathcal{C}$ then $ B \setminus A \in \mathcal{C}$ by $(ii)$
Now take $A_1, A_2 , \ldots A_n, \ldots$ an increasing sequence $A_i \subset A_{i+1}$. Then $A = \cup_i A_i$ is in $\mathcal{C}$ by $(iii)$
This proves that $\mathcal{C}$ is a $\lambda$-system containing the $\pi$-system $\mathcal{A}$. By Dynkin $\pi-\lambda$ theorem (https://en.wikipedia.org/wiki/Dynkin_system) $\mathcal{C} = \sigma(\mathcal{A})$
So now we know that $1_{A_1}, \ldots, 1_{A_N}$ are in $\mathcal{H}$ for any finite choice of $A_i \in \sigma(\mathcal{A})$, by $(ii)$ $a_1 1_{A_1} + \ldots + a_N 1_{A_N}$ is in $\mathcal{H}$.
This is the claim of the author.
If you would like to see that any $\sigma(\mathcal{A})$-measurable function is in $\mathcal{H}$ It suffices to consider $f = f^+ - f^-$ and to prove that $f^+,f^-$ are in $\mathcal{H}$.
The argument is essentially the one you presented.
Let $f$ be $\sigma(\mathcal{A})$-measurable. define $$g_n =\sum_{k=0}^{n2^n -1} \frac{k}{2^n} 1_{\{\frac{k}{2^n} < f^+ \le \frac{k+1}{2^n}\}} + n 1_{f^+ > n}$$ where $f^+ = f \vee 0$. Then $g_n \in \mathcal{H}$ once $\{x: \frac{k}{2^n} < f^+ \le \frac{k+1}{2^n} \} \in \sigma(\mathcal{A})$ and $\{x: f^+ > n\} \in \sigma(\mathcal{A})$ (because $f$ is $\sigma(\mathcal{A})$-measurable). Then $g_n$ increasingly converges to $f^+$. Since $f$ is bounded so is $f^+$ and therefore by $(iii)$ $f^+ \in \mathcal{H}$. The argument for $f^{-}$ is analogous.
this concludes the proof.
Both results are actually equivalent. You can prove one from the other.
Regarding the first result:
Let $\mathcal{C}$ be a class of subsets of $\Omega$ closed under finite intersections and containing $\Omega$. Let $\mathcal{B}$ be the smallest class containing $\mathcal{C}$ which is closed under increasing limits and by difference. Then $\mathcal{B} = \sigma ( \mathcal{C})$.
Some books call it "Monotone Class Theorem", although this is not the most usual naming.
A class having $\Omega$, closed under increasing limits and by difference is called a "Dynkin $\lambda$ system". A non-empty class closed under finite intersections is called a "Dynkin $\pi$ system".
The result above can be divided in two results
1.a. A $\lambda$ system which is also a $\pi$ system is a $\sigma$-algebra.
1.b. Given a $\pi$ system, the smallest $\lambda$ system containing it is also a $\pi$ system.
Some books call result 1.a (or result 1.b) "Dynkin $\pi$-$\lambda$ Theorem.
Some quick references is
https://en.wikipedia.org/wiki/Dynkin_system
The second result
Let $G$ be an algebra of sets and define $M(G)$ to be the smallest monotone class containing $G$. Then $M(G)$ is precisely the $\sigma$-algebra generated by $G$, i.e. $\sigma(G) = M(G)$.
Where a monotone class in a set $R$ is a collection $M$ of subsets of $R$ which contains $R$ and is closed under countable monotone unions and intersections.
is usually called "Monotone Class Lemma" (or theorem) you can find it in books like Folland's Real Analysis or Halmos' Measure Theory. In fact, Halmos presents a version of this result for $\sigma$-rings.
Let $G$ be ring of sets and define $M(G)$ to be the smallest monotone class containing $G$. Then $M(G)$ is precisely the $\sigma$-ring generated by $G$.
Let us prove that the results are equivalent
Result 1: Let $\mathcal{C}$ be a class of subsets of $\Omega$ closed under finite intersections and containing $\Omega$. Let $L(\mathcal{C})$ be the smallest class containing $\mathcal{C}$ which is closed under increasing limits and by difference. Then $L(\mathcal{C}) = \sigma ( \mathcal{C})$.
Result 2: Let $G$ be an algebra of sets and define $M(G)$ to be the smallest monotone class containing $G$. Then $M(G)$ is precisely the $\sigma$-algebra generated by $G$, i.e. $\sigma(G) = M(G)$.
Where a monotone class in a set $R$ is a collection $M$ of subsets of $R$ which contains $R$ and is closed under countable monotone unions and intersections.
Proof:
(2 $\Rightarrow$ 1). Note that any class containing $\mathcal{C}$ which is closed under increasing limits and by difference is close by complement because $\Omega \in \mathcal{C}$, and so it is also closed by decreasing limits. So it is closed under countable monotone unions and intersections. It means: any class containing $\mathcal{C}$ which is closed under increasing limits and by difference is a monotone class.
Note also that any class containing $\mathcal{C}$ which is closed under increasing limits and by difference contains $A(\mathcal{C})$ the algebra generated by $\mathcal{C}$.
Then using Result 2 we have
$$ \sigma(\mathcal{C}) = \sigma(A(\mathcal{C})) = M(A(\mathcal{C})) \subseteq L(A(\mathcal{C}))=L(\mathcal{C}) $$
Since $\sigma(\mathcal{C})$ is a class containing $\mathcal{C}$ which is closed under increasing limits and by difference, we have $L(\mathcal{C}) \subseteq \sigma(\mathcal{C})$, so $L(\mathcal{C}) = \sigma(\mathcal{C})$.
(1 $\Rightarrow$ 2). First let us prove that $M(G)$ is a class containing $G$ which is closed under increasing limits and by difference. Since $M(G)$ is monotone, we have that $M(G)$ is closed under increasing limits.
Now, for each $E\in M(G)$, define
$$M_E=\{ F \in M(G) : E\setminus F , F \setminus E \in M(G) \}$$
Since $M(G)$ is a monotone class, $M_E$ is a monotone class. Moreover, if $E\in G$ then for all $F \in G$, $F\in M_E$, because $G$ is an algebra. So, if $E\in G$, $G \subset M_E$. So, if $E\in G$, $M(G) \subset M_E$. It means that for all $E\in G$, and all $F \in M(G)$, $F \in M_E$. So, for all $E\in G$, and all $F \in M(G)$, $E \in M_F$. So, for all $F \in M(G)$, $G \subset M_F$, but since
$M_F$ is a monotone class, we have, for all $F \in M(G)$, $M(G)\subset M_F$. But that means that $M(G)$ is closed by differences.
So we proved that $M(G)$ is a class containing $G$ which is closed under increasing limits and by difference.
So by Result 1, $$\sigma(G)=L(G) \subseteq M(G)$$
Since $\sigma(G)$ is a monotone class, we have
$$ M(G) \subseteq \sigma(G)$$
So we have $$\sigma(G)= M(G)$$
Best Answer
Real Analysis by Stein and Shakarchi has a proof that avoids the monotone class theorem and Dykin's $\pi$-$\lambda$ theorem; I would not be surprised if this proof is Fubini's original proof. See Chapter 2 Section 3 for the proof for Lebesgue measure. See Chapter 6 Section 3 for the proof for general measures. The proof in the general case is indeed a generalization of the proof in the Lebesgue case. Introduction to Real Analysis by Christopher Heil gives essentially the same proof as Stein and Shakarchi, in the case of Lebesgue measure.
The proof is long, so I will only give a sketch here. By the usual linearity and convergence arguments, it suffices to prove the theorem for indicator (i.e., characteristic) functions of sets $E$. The idea is to first prove the theorem for measurable rectangles $E$, then for sets $E$ in $\mathcal{A}$ (where $\mathcal{A}$ is the algebra of finite disjoint unions of measurable rectangles), then for sets $E$ in $\mathcal{A}_{\sigma}$ (countable unions of sets in $\mathcal{A}$), then for sets $E$ in $\mathcal{A}_{\sigma \delta}$ (countable intersections of sets in $\mathcal{A}_{\sigma}$). Finally, to prove the theorem for arbitrary measurable sets, we use the following proposition whose proof is implicit in the Caratheodory construction.
Proposition 1.6 (Chapter 6, Section 1). Let $\mu_{*}$ be the outer measure generated by a premeasure $\mu_0$ on an algebra $\mathcal{A}$ on a set $X$. For any set $E \subseteq X$ and any $\epsilon>0$, there is an $E_1 \in \mathcal{A}_{\sigma}$ and an $E_2 \in \mathcal{A}_{\sigma \delta}$ such that $E \subseteq E_2 \subseteq E_1$ and $$\mu_{*}(E_1) - \epsilon \leq \mu_{*}(E) = \mu_{*}(E_2).$$
Note: I have not carefully verified the correctness of Stein and Shakarchi's proof of Fubini's theorem.