Prove: if a $\sigma$-algebra over $X$ contains one unique atom $A$, then $A=X$. Let $\mathcal{F}$ be a $\sigma$-algebra over some non-empty $X$. Given that $A$ is the only atom of $\mathcal{F}$ (recall $A \in \mathcal{F}$ is an atom of $\mathcal{F}$ if it's non-empty and no proper subset of $A$ other than the empty set is in $\mathcal{F}$),  show that $A=X$.
My attempt: It's clear $A \subset X$. To show $X \subset A$, suppose for contradiction there's some $x \in X$ and $x \in A^c$. Let
$$
B = \bigcap_{S \in \mathcal{F}: x \in S}  S
$$
(which is clearly non-empty as $x \in A^c \in \mathcal{F}$). My idea is to show that $B$ is also an atom of $\mathcal{F}$ (basically using this idea), and $A\cap B = \emptyset \implies A \neq B$, contradicting that $A$ is the unique atom. However, I'm stuck on showing $B \in \mathcal{F}$, because the intersection defining $B$ might not be countable. It's intuitively clear that if $\mathcal{F}$ has $n \in \mathbb{N}$ distinct atoms, then $ | \mathcal{F}| = 2^n$ (or at least $\mathcal{F}$ should be finite), but I have yet to prove this (it's actually the last part of this problem, and I'm stuck on part 1)...
 A: As I understand it, the problem as stated is false.
Let $Y$ be a set with an atomless $\sigma$-algebra $\mathcal{G}$ (see here for examples). Let $X = Y \sqcup \{a\}$ for some $a$. Define a $\sigma$-algebra $\mathcal{F}$ on $X$ by $E \in \mathcal{F}$ if and only if $E \backslash \{a\} \in \mathcal{G}$. In other words, $E \in \mathcal{F}$ exactly when $E \in \mathcal{G}$ or $E$ is the union of a set in $\mathcal{G}$ with $\{a\}$.
In this scenario, we have that $\{a\}$ is an atom. On the other hand, every other nonempty set $E \in \mathcal{F}$ satisfies $E \backslash \{a\} \neq \emptyset$, and since $\mathcal{G}$ is atomless we can find $\emptyset \neq F \subsetneq E$ with $F \in \mathcal{G}$, which also means $F \in \mathcal{F}$. Thus, $E$ contains a proper nonempty subset in $\mathcal{F}$, so is not an atom.
As noted in the comments below, the problem works out if the $\sigma$-algebra is countably generated. In this case, we can follow the idea in the question, making just one modification.
Suppose $\mathcal{F}$ is generated by some countable collection $\mathcal{E}$. We might as well assume $\mathcal{E}$ is an algebra, since the algebra generated by a countable collection is still countable. With this in hand, we can restrict our intersection defining B to sets $S \in \mathcal{E}$, and B should still be an atom.
