Subset of a countable set is itself countable How is it proved that if $$A \subset B\ \text{with}\ B\ \text{countable} $$ then $A$ is either countable, finite, or empty? I think the proof involves a $1-1$ correspondence between $\mathbb{N}$ and $A$ but other than that I do not know how to proceed.
EDIT: I have checked the solution and it advises me to proceed as follows. " As a start to a definition of $$g: \mathbb{N} \rightarrow A$$ set $$g(1)=f(n_1) $$ Then show how to inductively continue this process to produce a $1-1$ function $g$ from $\mathbb{N}$ onto $A$."
So the proof according to my book involves induction.
 A: 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 $f(1),f(2),...,f(n-1)$.
A: 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 $A$?
A: 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, $|A|\leq\aleph_0$.
A: We start with a bijective mapping
$\tag 1 f: \Bbb N \cong B$
Using $f$ we define another function
$\tag 2 \Phi: \mathcal P(B) \setminus \emptyset \to B$
$\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad X \mapsto f\bigr(\,\text{min}(f^{-1}(X))\,\bigr)$
If $A \subset B$ and $A$ is not finite, we can recursively build another function $g: \Bbb N \to A$ by setting
$\quad g(1) = \Phi(A)$
and with $g(1), g(2), \dots, g(n)$ set, defining
$\quad g(n+1) = \Phi\bigr(A \setminus \{g(1), g(2), \dots, g(n)\}\bigr)$
Exercise 1: Show that $g$ is a well-defined (recursive) function.
Exercise 2: Show that $g$ is a bijective correspondence,
$\tag 3 g: \Bbb N \cong A$

Here is another proof. 
A set $C$ is countable if there exist a surjection $h: \Bbb N \to C$. In fact,  an explicit bijection $\Bbb N \cong C$ can be constructed using $h$ (see, for example, this).
Addressing the interesting situation, let $B$ be an infinite set contained in a countable set $A$. Let $f: \Bbb N \cong A$ exhibit the $\text{1:1}$ correspondence. Choose any element $b_0 \in B$ and define
$$
F(m) = \left\{\begin{array}{lr}
        f(m), & \text{for } m \in f^{-1}(B) \\
       b_0 , & \text{for } m \in \mathbb N \setminus  f^{-1}(B)
        \end{array}\right\}
$$
The mapping $F: \Bbb N \to B$ is a surjection.
A: Hint: Suppose $f$ is the function that maps $B$ to $\mathbb{N}$, sort each item in $B$ by it's $f$ value.
