How can we prove that there is a finite set for every infinite set, in an infinite list of infinite sets. Given an infinite list of infinite sets of natural numbers (each set is sorted by size of the elements), I would like to find a proof that says we can always find a unique finite set of x natural numbers that correspond to the first x natural numbers in each infinite set (see example below).
I think that a proof is possible since we always have an infinite number of digits to choose from, but we will always need only a finite number of digits for our finite set.
I just don't know how to write such a proof, any help in how to approach this would be appreciated. Or, if this is not possible, how could we prove the opposite.

$1 \to \{ \color{red}{1} ,2,3,4,5,6,7,8 \dots \} \mapsto \{ \color{red}{1} \}$
$2 \to \{ \color{red}{2} ,4,6,8,10,12,14 \dots \} \mapsto \{ \color{red}{2} \}$
$3 \to \{ \color{red}{1,3} ,5,7,9,11,13,15 \dots \} \mapsto \{ \color{red}{1,3} \}$
$4 \to \{ \color{red}{1,2} ,3,7,9,19,27,31 \dots \} \mapsto \{ \color{red}{1,2} \}$
$5 \to \{ \color{red}{1,2,3} ,4,21,22,25,32 \dots \} \mapsto \{ \color{red}{1,2,3} \}$
$6 \to \{ \color{red}{2,3} ,4,6,7,8,21,55,58 \dots \} \mapsto \{ \color{red}{2,3} \}$
$7 \to \{ \color{red}{2,3,4} ,6,7,8,9,21,55,58 \dots \} \mapsto \{ \color{red}{2,3,4} \}$
$\dots$

 A: Taking TonyK's interpretation of the question:

I think the OP wants to know whether their idea works: given a sequence $(S_n)$ of infinite subsets of $N$, we can index them using finite subsets, by taking the $n$th index set to be the shortest initial segment of $S_n$ that has not yet occurred as an index set.

Yes, this does work.
The fact that makes it work is that at each step in the construction there are infinitely many different finite prefixes of $S_n$ to choose from, but only finitely many of these can have been used for earlier sets yet. So there is always some unused prefix to choose, and therefore a shortest unused prefix.
The above paragraph would count as a full proof in everyday mathematics.
If you want to reduce it all the way down to the nuts and bolts of axiomatic set theory, you'd be looking for a proof that the above recursive description actually defines a function with domain $\mathbb N$. For this you should look for an appropriate variant of the recursion theorem.
A: 
Well-ordering principle: Every non-empty set of natural numbers contains a (unique) least element.

Let $X\subseteq\mathbb{N}$ be an infinite set of natural numbers. Then for any $n\in\mathbb{N}$ there exists a unique finite set $F_n\subseteq X$ containing the $n$ least elements of $X$.
Proof: By induction on $n$. For the basis case $n=1$, the well-ordering principle guarantees a least element $x\in X$. Take $F_1=\{x\}$ and we are done.
For the inductive step, suppose we have a finite set $F_n$ containing the $n$ least elements of $X$, and we wish to form a new finite set $F_{n+1}$ containing the $n+1$ least elements. Consider the set $X\setminus F_n$, the elements of $X$ which are not in $F_n$. Since $X$ is infinite and $F_n$ is finite, $X\setminus F_n$ must be infinite, thus it is non-empty. By the well-ordering principle, $X\setminus F_n$ must contain a least element, call it $x$. Let $F_{n+1}=F_n\cup\{x\}$, so that $F_{n+1}$ is the set of $n+1$ least elements of $X$, completing the proof.
Comments: The proof should be intuitive. To form the finite set, keep selecting the least element from the infinite set, guaranteed by well-ordering principle, subtract that element, and repeat until you have a set of the proper size. 
