Let $\mathcal{F} \subset \mathcal{P}{(\Omega)}$. Then $\mathcal{F}$ is a $\sigma$-algebra if and only if $\mathcal{F}$ is an algebra and satisfies $A_n \in \mathcal{F}, A_n \subset A_{n+1}$ for all $n \implies \bigcup_{n\geq 1} A_n \in \mathcal{F}$.
Proof given:
I don't get the logic of the 'if' part. To prove that F is a sigma algebra shouldn't we refer back to the definition and prove that it satisfies?
Since $\mathcal{F}$ is already an algebra, we only need to show that if $\{B_n\}_{n\in \mathbb{N}}\subset \mathcal{F}$ , then $$\bigcup^\infty_{n=1} B_n \in \mathcal{F},$$ that is, we need to show that $\mathcal{F}$ is closed under countable unions.
So, if $\{B_n\}_{n\in \mathbb{N}}\subset \mathcal{F}$, define $$A_n = \bigcup^n_{k=1} B_k.$$ Notice that $A_n \in \mathcal{F}$, because $\mathcal{F}$ is an algebra and $A_n\subset A_{n+1}$.
Hence, by the 'only if' hipothesys, we know that $$\bigcup^\infty_{n=1}A_n = \bigcup^\infty_{n=1} \bigcup^n_{k=1} B_k = \bigcup^\infty_{k=1} B_k \in \mathcal{F} $$ showing that $\mathcal{F}$ is indeed a $\sigma$-algebra.