A problem on smallest $\sigma-$ algebra in Axler's Measure theory book Could you please give me some hints on this problem please (this problem is from Measure, Integration and Real Analysis, section 2.28, Axler, 2020)

$X$ is a set. $A$ is the set of subset of $X$ such that $A$ = {{$x$}: $x \in X$}.
Prove that the smallest $\sigma-$ algebra on $X$ that contains $A$ is the set of all subset $E$ of $X$ such that $E$ is countable or $X$\ $E$ is countable (which I will call the set $B$ from now on).

My idea is to prove that $B$ is the intersection of all $\sigma-$algebra on $X$ that contains $A$. To do it, I must prove that every element of $B$ $\in$ an arbitrary $\sigma-$ algebra on X that contains $A$.
But from here, I'm stuck at how to prove a set is in a $\sigma-$ algebra.
Thank you very much for any comments or any hints.
 A: 
$X$ is a set. $A$ is the set of subset of $X$ such that $A$ = {{$x$}: $x \in X$}.
Prove that the smallest $\sigma-$ algebra on $X$ that contains $A$ is the set of all subset $E$ of $X$ such that $E$ is countable or $X$\ $E$ is countable (which I will call the set $B$ from now on).

Proof: Let us write $\sigma(A)$  to indicate the smallest $\sigma-$ algebra on $X$ that contains $A$. Let $B$ be the set of all subset $E$ of $X$ such that $E$ is countable or $X \setminus E$ is countable
Item a. In this exercise, it is need to prove that $B$ is, in fact, a $\sigma$-algebra. It can be done by just checking the definition.
$1$. $\emptyset$ is countable, so $\emptyset \in B$.
$2$. If $E \in B$ we have two possibilities: $E$ is countable or $X$\ $E$ is countable.
$2$.a.If $E$ is countable, then $X \setminus E^c= X \setminus (X \setminus E)=E$ is countable, so $E^c\in B$.
$2$.b. If $X \setminus E$ is countable, then $ E^c = X \setminus E$ is countable, so $E^c\in B$.
So, in both cases, we have that $E^c\in B$.
$3$. Given a countable collection $\{ E_n\}_n$ such that, for all $n$, $E_n \in B$, we have that:
$3$.a If for all $n$, $E_n$ is countable, then $\bigcup_n E_n$ is countable and so, $\bigcup_n E_n \in B$.
$3$.b If there is one $n_0$ such that $X \setminus E_{n_0}$ is countable, then
$X \setminus \bigcup_n E_n= \bigcap (X\setminus E_n) \subseteq X \setminus E_{n_0}$ is countable and so, $\bigcup_n E_n \in B$.
So, in both cases, we have that $\bigcup_n E_n \in B$.
So $B$ is a $\sigma$-algebra.
Item b. Let us now prove that $\sigma(A) \subseteq B$.
Since for each $x \in X$ , $\{x\}$ is countable, we have that $ A \subseteq B$. Since $B$ is a $\sigma$-algebra, it follows that $$ \sigma(A) \subseteq B \tag{1}$$
Item c. Now let us prove that $ B \subseteq \sigma(A) $.
Let $E \in B$.
If $E$ is countable, then $E$ can be written as a countable union of single-point subsets (that is $E=\bigcup_{x \in E} \{x\}$). Since all single-point subsets are in $A$ and $\sigma(A)$ is a $\sigma$-algebra, we have that $E \in \sigma(A) $.
If $X \setminus E$ is countable, then, by the previous paragraph, $X \setminus E \in \sigma(A) $. Since $\sigma(A)$ is a $\sigma$-algebra, we have that $E \in \sigma(A) $.
So, in both cases, $E \in \sigma(A) $. So, $$ B \subseteq \sigma(A)  \tag{2}$$
From $(1)$ and $(2)$, we have $\sigma(A) = B$.
