Equivalence of two countable-definitions This is an equivalence proof by myself. I need someone to tell me if the proof is alright or to point out its flaws.
Definition 1: A set M is countable iff $|M| = |\Bbb N|$ or |n $\in \Bbb N $| = |m $\in M$| (here I mean that if M is finite then you can just pair the element m $\in$ M to a natural number n $\in \Bbb N$ and it will become a finite bijection between all the elements of M and the same amount of numbers of $\Bbb N$).
Definition 2: A set M is countable iff $|M| \le |\Bbb N|$, i.e. there is some injective $g: M \to \Bbb N$.
So now I want to prove equivalence of the two definitions:
1 $\to$ 2: This is easy because if $|M| = |\Bbb N|$ or |n $\in \Bbb N $| = |m $\in M$|then also always $|M| \le |\Bbb N|$.
2 $\to$ 1: I split in two cases.
Case 1: $|M| \le |\Bbb N|$, i.e. there is some injective $g: M \to \Bbb N$, and rng(g) is finite. Then this leads naturally to the second variant of definition 1 because we can just take the n elements of rng(g) and map it into M which must be bijective.
Case 2: $|M| \le |\Bbb N|$, i.e. there is some injective $g: M \to \Bbb N$, and rng(g) is infinite. So it must be |rng(g)| = |$\Bbb N$| = |M| which lets us arrive at variant 1 of definition 1.
But why is |rng(g)| = |$\Bbb N$|? Because rng(g) $\subseteq \Bbb N$ which means that rng(g) is well-ordered as well and so every non-empty subset of rng(g) has a minimum element. So you can construct the following chain:
a_minimum $\in$ rng(g)
b_minimum $ \in$ rng(g)\a_minimum
c_minimum $ \in$ rng(g)\a_minimum, b_minimum
...
Since rng(g) is infinite this goes on endless and you can cleary see the structure of the natural numbers rising, so that there's a bijection possible between rng(g) $\to \Bbb N$, so that |rng(g)|=|$\Bbb N$|.
But why |rng(g)| = |M|? Because of the definition of rng(g) we get that the inverse function rng(g) $\to$ M must be surjective but the function rng(g) $\to$ M is injective by assumption of definition 2: $|M| \le |\Bbb N|$. That's why |rng(g)| = |M|.
 A: Let $g(M)=H_0$ be an infinite subset of $\mathbb N$. You may define $H_i$ for $i=1,2,\dots$ recursively:

*

*$H_{i-1}$ is infinite non-empty by hypothesis, so by the well-ordering property of $\mathbb N$ it must have a minimal element, say $h_{i-1}$.

*Let $H_i = H_{i-1} \setminus\{h_{i-1}\}$.

*If $H_i$ was finite, then $H_{i} \in H_i \cup \{ h_{i-1} \}$ would be a union of two finite sets so a finite set, a contradiction. Thus $H_i$ is infinite.

*Observe $h_{i-1} \notin H_i$ and $H_i\subset H_{i-1}$.

Observe $H_0 \supset H_1 \supset H_2 \supset \cdots$. We will show in a moment that $\{h_i\}$ are all distinct.
Let $h: \mathbb N\to g(M)$ be defined by $h(n) = h_n$. Observe it is an injective map, otherwise $\exists i>j$ such $h(i) = h(j)$ i.e. $h_i = h_j$. Then
$$h_j=h_{i} \in H_i \subset H_{i-1} \subset \cdots  \subset H_{j+1}$$
but $h_{j} \in H_{j+1}$ which contradicts our observations. Thus $h$ must be injective and we have $|\mathbb N|\leq |g(M)|$. Clearly if we restrict $g$ to be $g:M\to g(M)$ it is a bijection so $|M| = |g(M)| \leq |\mathbb N|$ Is is clear from here?
Edit: added working with the image of $M$.
Edit: some typos.
Edit: format.
