set theory - prove that $S \subseteq \mathbb N^{\mathbb N} $ is Countable set

Question

Hey I have to submit an assignment to the University, I would really appreciate your thoughts on the proof I wrote.

Let $S \subseteq \mathbb N^{\mathbb N} $ :

$f \in S \iff \exists n \in \mathbb N ,\forall k>n \rightarrow f(k)=0$

prove that S is countable.

---

let there be $g:S \rightarrow \mathbb N×....×\mathbb N$

so that $~~g(f) = ~<f(0),f(1),f(2)...,f(n)>$ so that $~\forall k>n f(k)=0$

Let $~~f,h \in S ~~ $assume that $~~g(f)=g(h)$ we will show that $~~f=h$

$g(f) = <f(0),..f(n)>$

$g(h) = <h(0),...h(m)>$

so $f(0)=h(0), ... ,f(m)=h(n)$

$m=n~$ by a set definition and $~\forall k>m,n ~ f(k)=0=h(k)$

so $\forall x \in \mathbb N ~ f(x)=h(x)~$ and therefore $f=h$

so $|S| \leq |\mathbb N×....×\mathbb N| = |\mathbb N |$ and therefore S is countable.

Answer 1

For all n in N, show
S(n) = { f in S : for all k > n, f(k) = 0 } is countable.
Conclude S = $\cup_n$S(n) is a countable union
of countable sets, hence countable.

This is the same proof used to show there are countably many
polynomials with interger, or with rational, coefficients.