I found different statements of the Monotone Class Theorem. On probability Essentials (Jean Jacod and Philip Protter) the Monotone Class Theorem (Theorem 6.2, page 36) is stated as follows:
Let $\mathcal{C}$ be a class of subsets of $\Omega$ 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 $\mathcal{B} = \sigma ( \mathcal{C})$.
While on Wikipedia (https://en.wikipedia.org/wiki/Monotone_class_theorem) the theorem is:
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.
It looks like the second theorem should be a special case of the first. Does the first prove the second? Is it possible to prove the first from the second? Is there a decent literature on those two theorems?
Both results are actually equivalent. You can prove one from the other.
Regarding the first result:
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
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 us prove that the results are equivalent
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 $$\sigma(G)= M(G)$$