Show that any enumerable set is either finite or equinumerous with the natural numbers. I really do not know how to show this in any sound manner. For any enumerable set A there exists a surjective function f from N to A.
I then proceeded by arguing that if f is an injection it is a bijection and A would be infinite. Otherwise if f is not an injection f^-1 would still be total and therefore for every a ∈ A there exists a nonempty set M_a ⊂ N with f(m) = a for all m ∈ M_a. If we now construct a function g which assisgns to every a the smallest m ∈ M_a, then g is a bijection. Because g is a function mapping A to a subset of N, A would be finite.
But this does not hold because there are infinite subsets of N.
If anybody could tell me how to prove this id be very gratefull.
EDIT: enumerbility is defined as there being a surjective function from N to A
 A: Your constructing bijection $f$ from $A$ to $N_A\subseteq\mathbb N$ can certainly serve as a starting point. Now either $N_A$ is bounded  or it isn't. 
If it has there's an $n\in\mathbb N$ such that $j\in N_A\Rightarrow j<n$. This means that $N_A$ is finite (it's no larger than $n$).
If it hasn't we construct an surjection from $N_A$ to $\mathbb N$ which would show that $|N_A| \ge|\mathbb N|$ and since $N_A\subseteq\mathbb N$ it would prove that $N_A$ is equinumerous to $\mathbb N$. Now define $G_n = \{x\in N_A: x < n\}$, now $g(n) = |G_n|$ is such a mapping. 
The latest claim can be shown by induction. Since $g(0)=0$ we have that $0$ is part of the range of $g$. And if $j$ is in the range, that is there's an $n$ such that $j = g(n)$ then we can just take the minimum of $N_A\setminus G_N$ to find a $m$ such that $g(m) = j+1$.
Now we have $f:A\to N_A$ bijectively and $g:N_A\to \mathbb N$ bijectively, this means that $g\circ f: A\to\mathbb N$ bijectively which means that $A$ and $\mathbb N$ are equinumerous.
