Subset of a countable set is itself countable

Any subset of a countable set is countable.

Take $A\subset B$ where $B$ is countable. Then $|A|\leq|B|$ since $A\subset B$. By definition, $|A|\leq|B|$ if there is a one-to-one function from $A$ into $B$. We also see that $|B|\leq\aleph_0$ since $B$ is countable. Thus, $|A|\leq\aleph_0$.

Via the bijection $\Bbb N\cong B$ we have an injection $i:A\to\Bbb N$.

Define $$f(n+1)=\min\{k∈i[A]∣k>f(n)\}$$ for $n≥1$ and $$f(1)=\min i[A]$$

We claim that for each $n$ we have $\{f(1)<f(2)<...<f(n)\}$ is a subset of $i[A]$ and contains each $i(a)$ less than $f(n)$.
Proof: Induction over $n$:
For $n=1$ clearly $f(1)∈i[A]$, and since there is no $i(a)<f(1)$, the claim is true.
Assume that the statement is true for $n$. Then by definition of $f$, the number $f(n+1)$ is larger than $f(n)$, so the set $\{f(1)<...<f(n)<f(n+1)\}$ is a subset of $i[A]$. Since $f(n+1)$ is the minimal element of $i[A]$ larger than $f(n)$, it must contain each $i(a)$ less than $f(n+1)$.

So we have shown that $f:\Bbb N↦i[A]$ is an injection. Now, for $a\in A$ we have the natural number $l=i(a)$. Since $l$ is less than $f(n)$ for some $n\in\Bbb N$, $f$ being strictly increasing, it must thus be one of $f(1),f(2),...,f(n-1)$.

Hint: Define an injection $f:B \to \mathbb{N}$ (this is possible as B is countable). Define the inclusion mapping $I:A \to B$. Consider $f\circ I:A \to \mathbb{N}.$ What can you say about $I$? What can you then say about $f \circ I$? What can you then conclude about $A$?