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
In order to turn any sequence of sets into a monotone increasing or decreasing sequence of sets, you have to use the finite unions/finite intersections (respectively) and set complements.
A set algebra is precisely a collection of sets which is closed under finite unions, finite intersections, and set complements.
Note also that being closed under finite unions and set complements implies being closed under finite intersections, and being closed under finite intersections and set complements implies being closed under finite unions, by DeMorgan's Laws.
I'm trying to find a source for this, and I can't right now, but I believe if you have a countable union of sets, not necessarily increasing, i.e. $$B = \bigcup\limits_{i=1}^{\infty} B_i$$ then you can always write it as the limit of an increasing union of sets by setting: $$A_1=B_1, A_2= B_1 \cup B_2, \dots, A_n = B_1 \cup \dots \cup B_n $$ Then clearly $\bigcup_{i=0}^{\infty} B_i=\bigcup_{i=1}^{\infty} A_i$. Since each of the $A_i$ is defined using only a finite union, it is guaranteed to be in the system of sets if that system is a set algebra. Then because it is a monotone class, it contains the union of the $A_i$, thus the union of the $B_i$ which is equal, and therefore it is even a $\sigma-$algebra, since $B_i$ was an arbitrary countable family.
If you want to write in the form $\lim_{i \to \infty} C_i = \emptyset$, then it suffices to write $$C_n = A_n \setminus A_{n-1} $$ and we are only guaranteed that each $C_i$ is in our set system if it is a set algebra because otherwise the set system may not be closed under set complements.
And of course $$\bigcup_{i=0}^{\infty} B_i=\bigcup_{i=1}^{\infty} A_i = \bigcup_{i=1}^{\infty} C_i$$
The procedure for showing closure under countable intersections is equivalent; just use the fact that your set system is closed under set complements combined with DeMorgan's Laws.