Proving that $\mathcal{S} = \{E\subset X: E \text{ is countable or } X\setminus E \text{ is countable}\}$ is the smallest $\sigma$-algebra 
Suppose $X$ is a set and $\mathcal{A}$ is the set of subsets of $X$ that consist of exactly one element: $$\mathcal{A} = \{\{x\}:x \in X\}.$$ Prove that the smallest $\sigma$-algebra on $X$ containing $\mathcal{A}$ is the set of all subsets $E$ of $X$ such that $E$ is countable or $E^c$ is countable.

To prove this, I first want to show that $\mathcal{S}$ is a $\sigma$-algebra where $$\mathcal{S} = \{E\subset X: E \text{ is countable or } X\setminus E \text{ is countable}\}.$$
My question: I've already shown that $\mathcal{S}$ is a $\sigma$-algebra. Now, I think I need to show that $\mathcal{A}$ is contained in $\mathcal{S}$. How can I show this (since both $\mathcal{A}$ and $\mathcal{S}$ are sets of sets)? So, how can $\mathcal{A}$ be contained in $\mathcal{S}$? Also, lastly, I think I need to show that $\mathcal{S}$ is the intersection of all $\sigma$-algebras on $X$ that contain $\mathcal{A}$. Can someone show how this can be proven?
 A: As you said you have already proved that
$$\mathcal{S} = \{E\subseteq X: E \text{ is countable or } X\setminus E \text{ is countable}\}$$
is in fact a $\sigma$-algebra.
Now, we have $\mathcal{A} = \{\{x\}:x \in X\}$ and let us prove that the smallest $\sigma$-algebra containing $\mathcal{A}$  is equal to $\mathcal{S}$.
Since for each $x \in X$ , $\{x\}$ is countable, we have that $ \mathcal{A} \subseteq \mathcal{S}$.
Since $\mathcal{S}$ is a $\sigma$-algebra, it follows that
$$ \sigma(\mathcal{A}) \subseteq \mathcal{S} \tag{1}$$
where $\sigma(\mathcal{A})$ is the smallest $\sigma$-algebra containing $\mathcal{A}$ (the $\sigma$-algebra generated by $\mathcal{A}$).
Now let us prove that $ \mathcal{S} \subseteq \sigma(\mathcal{A}) $.
We will begin by proving the claim:

Every countable set $F\subseteq X$ is in $\sigma(\mathcal{A})$.

Proof of the claim:
Since for every $x \in X$ , $\{x\} \in \mathcal{A} \subseteq \sigma(\mathcal{A})$ and every countable set $F\subseteq X$ is a countable union of single-point sets, we have that $F \in \sigma(\mathcal{A})$ (because $\sigma(\mathcal{A})$ is a $\sigma$-algebra). So, every countable subset of $X$ is in $\sigma(\mathcal{A})$ . $\square$
Now, for any $E \in \mathcal{S}$, we have two possibilities:

*

*$E$ is countable. In this case, it follows directly from the claim that $E \in \sigma(\mathcal{A})$.

*$X\setminus E$ is countable. In this case,  it follows directly from the claim that $X\setminus E \in \sigma(\mathcal{A})$. Since  $\sigma(\mathcal{A})$ is a $\sigma$-algebra, we have $X\setminus(X\setminus E) \in \sigma(\mathcal{A})$. But $X\setminus(X\setminus E) =E$, so we have $E \in \sigma(\mathcal{A})$.

So, in both cases, we have that  $E \in \sigma(\mathcal{A})$.
So, we have:
$$ \mathcal{S}= \{E\subseteq X: E \text{ is countable or } X\setminus E \text{ is countable}\} \subseteq \sigma(\mathcal{A})  \tag{2}$$
From $(1)$ and $(2)$,
$$ \sigma(\mathcal{A}) = \mathcal{S} $$
It means, the smallest $\sigma$-algebra containing $\mathcal{A}$  is equal to $\mathcal{S}$.
