It can be shown that any infinite sigma algebra has cardinality greater than or equal to that of the continuum.
This is not immediate. I don't think it works quite that simply, there is no guarantee that two sequences will give different sets.
You probably need to use the axiom of choice. Let X be a set and an infinite sigma algebra over X. Then define the equivalence relation ~ as x~y if and only if . Then each equivalence class E(x) is measurable, as it is the countable intersection of sets in the sigma-algebra containing x.
Let I be the set of representatives from each equivalence class (Axiom of Choice here). Then is a disjoint union. Hence if we suppose that is countable then I must either be finite or countable. Thus we have an injection from to as we can pick an element of by taking (as they are equivalence classes, you do have a guarantee of getting distinct sets).
This is impossible as if I is countable or finite.
I don’t understand your objection. If is a countable sequence of disjoint, nonempty sets, then using different index sets, the unions over the will be distinct sets. I.e., if where , then . Given an arbitrary countable sequence in a -algebra, you can of course make a new sequence of pairwise disjoint, nonempty sets (without AC).
I don’t think AC is necessary for the proof (c.f. mine above). Also, here is another proof which takes a different approach.