Does an element of a set, that can't be in a list, make that set uncountable? We can map the infinite subsets of $\mathbb N$ to the finite subsets of $\mathbb N$
The finite subset will be a prefix, of the infinite subset (that it is paired with), that has not yet been used.

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

We can then find an infinite set that is not in this list (by diagonalization).

$N \mapsto \{ \color{red}{4,5,8,9,\dots}   \} \mapsto \{ \color{red}{4} \} \lor \{ \color{red}{4,5} \} \lor \{ \color{red}{4,5,8} \} \lor\{ \color{red}{4,5,8,9} \} \dots$

We can then pair our infinite set, that can't be in our list, with $\{ \color{red}{4} \}$ if it is not been used or $\{ \color{red}{4,5} \}$  or $\{ \color{red}{4,5,8} \}$ etc... until we find some finite set that has not been used.
Since we need sets with a finite number of elements, and we have sets with an infinite number of elements to choose from, we will always find a finite prefix from each infinite set.
So, if we can find a finite set for every infinite set IN or NOT IN our list, and our set of finite sets is countable then can we still say that our set of infinite sets is uncountable based on there being an element that can't be in a list?
 A: Even though that specific infinite subset isn't in our list, we can still have all its finite prefixes in our list of finite subsets.
For instance, say the 10th infinite subset in your list is $\{4, \ldots\}$. Then the 20th is $\{4, 5, \ldots\}$. Then the 30th is $\{4, 5, 8, \ldots\}$. And so on. None of them have to be your infinite set, all of them can be different from your infinite set early enough to allow your set to be the result of the diagonalization. But all of your set's prefixes are in our list of finite subsets.
A: Your question is moot, since the motivating "fact" is false.

The culprit is your implicit claim in the sentence

until we find some finite set that has not been used.

How do we know such a thing exists at all?
For example, let $(F_i)_{i\in\mathbb{N}}$ enumerate the finite sets, and let $X_i=F_i\cup (\max(F_i),\infty)$ (I'm just taking the unions above in order to make each of my $X_i$s infinite). Note that $F_i$ is a prefix of $X_i$. I can map each $X_i$ to $F_i$, and this seems to work ... but now I've used up every finite set, and when I come to my next infinite set - say, the set of evens - I'm stuck.
Now you might argue that I built my map stupidly above, but the point is that the onus is on you to prove that there is a non-stupid way to "continue building forever." And indeed, one of the consequences of Cantor's theorem is that there isn't.
