Proving that $\mathbb Z$ with the finite-closed topology satisfies the second axiom of countability. In my general topology textbook there is the following exercise:

A topological space $(X,\tau)$ is said to satisfy the second axiom of countability if there exists a basis $B$ for $\tau$, where $B$ consists of only a countable number of sets.

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*Let $(X,\tau)$ be the set of all integers with the finite-closed topology. Does the space $(X,\tau)$ satisfies the second axiom of countability.


I made a proof for this but in my prove I made a mistake that I'll point out with the number (1).

My proof
Let $A_i=\{\mathbb Z\setminus B : \text{card}\  B=i\}$, for $B \subset \mathbb Z$. Then we have that $\tau= \bigcup \limits _{i=0} ^\infty A_i$. If every set $A_i$ is countable then we have that $\tau$ is countable as well.
Let's fix a value of $i$ and define the set $C_i=\{X: \text{card} \ X=i\}$, for $X \subset \mathbb Z$.
Each $X$ is countable because it's finite, and because $C_i=\bigcup X$ (1), then $C_i$ is countable.
Let $f_i: C_i \to A_i$, with $f_i(X)=Z\setminus X$, then $f_i$ is bijective, thus $C_i \sim A_i$, thus $A_i$ is countable, proving that $\tau$ is also countable.
If $B$ is a basis for this space, then $B \subset \tau$, any subset of a countable set is also countable thus $B$ is countable, so $(\mathbb Z, \tau)$ does satisfy the axiom.

My mistake was that $C_i \neq \bigcup X$, but instead $C_i = \bigcup \{X\}$. How can I prove that $C_i$ is countable thus concluding the proof?
 A: You are right that $C_i\neq \bigcup X$ and that rather $C_i \bigcup {X}$. I think it is easiest to prove $C_i$ is countable by induction on $i$. For $i=0$ this is trivial (in fact $C_0$ has only one element: the empty set), and it's still pretty trivial for $i=1$ (since $C_1$ is more or less $\mathbb{Z}$ itself). Now, supposing $C_n$ is countable, it is pretty easy to show $C_{n+1}$ is countable as well.
For instance, you can say that, since $C_n$ is countable and $\mathbb{Z}$ is countable, also $C_n\times \mathbb{Z}$ is countable, and hence $S:=\{(A,z):A\in C_n,z\in\mathbb{Z}\setminus A\}$ is countable, and we a surjective map from $S$ onto $C_{n+1}$ (sending $(A,z)$ to $A\cup \{z\}$.
Thus, $C_n$ is countable for every $n$.
By the way, a remark: you proved that every basis of this space is countable. This is correct, but you didn't need to show that. You only needed to show that there exists a countable basis of this space.
A: Any countable set with the finite-closed topology has countably many open sets so, in particular, it satisfies the second axiom of countability.
For every infinite set $X$, the set $P_{<\omega}(X)$ of all finite subsets of $X$ has the same cardinality as $X$. In the case the set $X$ is countable, the proof is easy. If $F$ is a finite subset of $X$, define
$$
f(F)=\prod_{k} p_k^{\chi_F(k)}
$$
where $(p_k)$ is the sequence of the prime numbers and $\chi_F(k)=1$ if $k\in F$, $\chi_F(k)=0$ if $k\notin F$.
Then the function $f\colon P_{<\omega}(X)\to\mathbb{N}$ is injective, proving the claim.
Adding the whole set to the set of finite subsets doesn't change the cardinality, so the set of closed subsets is countable and the topology is thus countable as well.
