As stated in the title the question is "Show that an infinite $\sigma$-algebra contains a infinite sequence of disjoint nonempty sets". I've seen this question asked a lot here, many with different answers. I'd like to know if my approach to it is correct.

Consider $\mathcal{A}$ an infinite $\sigma$-algebra on $X$ and define $$\mathfrak{C}=\{\mathcal{C}\subset \mathcal{A}-\{\emptyset\},\, \mathcal{C}\text{ is a disjoint collection of subsets} \}$$ Let show that there exists finite $\mathcal{C}\in\mathfrak{C}$ of arbitrarily large size. If this wasn't the case, take $N=\max\{|\mathcal{C}|,\,\mathcal{C}\in \mathfrak{C}\}$, there exists $\mathcal{C}_0$ such that $|\mathcal{C}_0|=N$ and therefore $\mathcal{C}_0=\{E_1,\dots,E_N\}$.

If $\bigcup_{n=1}^N E_n \neq X$ then $\mathcal{C}_0$ would not be maximal as $\mathcal{C}_0\cup \{X-\bigcup_{n=1}^N E_n\}$ would be a collection of $N+1$ disjoint subsets, therefore we have $\bigcup_{n=1}^N E_n = X$. Now consider $\mathcal{A}'$, the $\sigma$-algebra generated by $\mathcal{C}_0$, $\mathcal{A}'$ will be finite due to $\mathcal{C}_0$ being finite.

Since $\mathcal{A}$ is infinite there exists $E\in \mathcal{A}-\mathcal{A}'$. For some $n_0$ we must have $E_{n_0}\not\subset E$ and $E_{n_0}\cap E\neq \emptyset$, otherwise we would have that for all $n$ either $E_n\subset E$ or $E_n\cap E=\emptyset$ meaning $$E=\bigcup_{E_n\cap E\neq \emptyset} E_n\in \mathcal{A}'$$ Now, set $E'_{n_0}:=E_{n_0}\cap E\in \mathcal{A}-\{\emptyset\}$ and $E_{N+1}:=E_{n_0}-E'_{n_0}\in \mathcal{A}-\{\emptyset\}$. We have $E_{n_0}=E'_{n_0}\cup E_{N+1}$ where the union is disjoint, moreover $$\mathcal{C}:=\{E_1,\dots,E'_{n_0},\dots,E_N,E_{N+1}\}\in \mathfrak{C}$$ and $|\mathcal{C}|=N+1$. This contradicts the maximality of $N$, therefore there exists finite $\mathcal{C}\in\mathfrak{C}$ of arbitrarily large size.

EDIT: As detailed in the comments there was a mistake, I'll try to see if I can salvage part of this proof.

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    $\begingroup$ Every set in $\mathfrak C$ been finite doesn't imply existence of $N$ such that all set is less then $N$. $\endgroup$ – mihaild Apr 12 at 15:40
  • $\begingroup$ Woops, you're right. Still I feel I can apply my construction below to increase any finite set to arbitrary size, am I right? $\endgroup$ – Julio Cáceres Apr 12 at 15:50
  • $\begingroup$ To amend it partially order $\mathfrak{C}$ by inclusion and look at maximal elements. $\endgroup$ – user647486 Apr 12 at 15:54
  • $\begingroup$ Yeah, I was thinking about that. Although I'm not sure if the inclusion is the right partial order to use. $\endgroup$ – Julio Cáceres Apr 12 at 15:57
  • $\begingroup$ @JulioCáceres Well, just do it. You will be sure at the end. $\endgroup$ – user647486 Apr 12 at 16:04

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