Quick question: Does the sigma-algebra generated by a countable number of sets have a countable number of elements?
Given a countably infinite $\displaystyle \sigma$-algebra $\displaystyle \Sigma$ over an infinite set X, let $\displaystyle \{E_n\}_{n\in\mathbb{N}}$ be an enumeration of $\displaystyle \Sigma$. Wlog, we can take these to be disjoint. Then, for every binary sequence, we may produce another set in $\displaystyle \Sigma$. Namely, consider a sequence $\displaystyle b=b_1b_2b_3\ldots$, and let $\displaystyle F_b=\bigcup_{b_i=1}E_i$.
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 $\displaystyle \mathcal{F}$ an infinite sigma algebra over X. Then define the equivalence relation ~ as x~y if and only if $\displaystyle x \in A \in \mathcal{F} \iff y \in A$. 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 $\displaystyle X= \bigcup_{x \in I} E(x)$ is a disjoint union. Hence if we suppose that $\displaystyle \mathcal{F}$ is countable then I must either be finite or countable. Thus we have an injection from $\displaystyle \{0,1\}^I$ to $\displaystyle \mathcal{F}$ as we can pick an element of $\displaystyle \mathcal{F}$ by taking $\displaystyle \bigcup_{x\in S \subset I}E(x)$ (as they are equivalence classes, you do have a guarantee of getting distinct sets).
This is impossible as $\displaystyle |\{0,1\}^I|> |I|$ if I is countable or finite.
I don’t understand your objection. If $\displaystyle \{E_i\}_{i\in\mathbb{N}}$ is a countable sequence of disjoint, nonempty sets, then using different index sets, the unions over the $\displaystyle E_i$ will be distinct sets. I.e., if $\displaystyle A,B\subset\mathbb{N}$ where $\displaystyle A\neq B$, then $\displaystyle \bigcup_{i\in A}E_i\neq\bigcup_{i\in B}E_i$. Given an arbitrary countable sequence in a $\displaystyle \sigma$-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.