Infinite Sigma algebra contains infinite sequence of nonempty disjoint sets I realize this question has been answered but I wonder if this argument would work. 
Proof: 
If the infinite Sigma algebra X contains an infinite sequence of strictly nested nonempty sets, then we are done. 
Otherwise, take the set of all maximal strict nests in X, then the smallest element of these nests are disjoint from each other. If there are infinitely many then we are done. 
Otherwise, delete those small elements from each set in X and repeat the process above. Since the process can be repeated infinitely many times and the resulting smallest set in each nest form a disjoint set, the proof is finished. 
 A: EDIT: your argument does indeed work, although it could be written a bit clearer (e.g. justify why smallest elements of nests exist, and what you mean by "repeat the process infinitely many times" - this isn't hard and is a bit tedious, but it makes for a more complete proof).

Here's another nice way to attack the problem:
Call our $\sigma$-algebra $\mathbb{S}$. Say that a set $A$ is splitting if $A\in\mathbb{S}$ and $$\{A\cap B: B\in\mathbb{S}\}$$ is infinite. (Note that this is equivalent to $\{B: B\in\mathbb{S}, B\subseteq A\}$ being infinite.) That is, $A$ is splitting if it has infinitely many "pieces" in the $\sigma$-algebra. Note that most natural $\sigma$-algebras contain all singletons; for such a $\sigma$-algebra, "splitting" is the same as "infinite and in the $\sigma$-algebra." Note that since the $\sigma$-algebra is finite, the whole underlying set is splitting (finite $\sigma$-algebras, by contrast, have no splitting sets).
Now the crucial fact is the following:

$(*)\quad$ If $A\in\mathbb{S}$ is splitting, then $A$ can be partitioned as $A=B\sqcup C$ where $B, C\in\mathbb{S}$ are nonempty and $C$ is splitting.

(Of course this partition will not in general be unique, and of course both pieces can be splitting.) Note that if $(*)$ is true, we're done: we build a sequence of strictly decreasing sets in $\mathbb{S}$ $X=A_0\supsetneq A_1\supsetneq A_2\supsetneq ...$ (here "$X$" denotes the whole space), and so the sets $B_i=A_i\setminus A_{i+1}$ are disjoint elements of $\mathbb{S}$.
So can you prove $(*)$? HINT: first show that any splitting $A$ has a proper subset which is an element of $\mathbb{S}$; now show that if $D\cup E$ is splitting and $D, E\in\mathbb{S}$, then at least one of $D$ and $E$ is splitting.
