We can map the finite subsets of the power set of $\mathbb N$ to the infinite subsets like this.
$ \{1\} \to \{1, x_{(1,1)},x_{(1,2)},x_{(1,3)}, \dots\}$
where $x_{(n,n)}$ represents a random number, where each number is greater than the number on it's left side.
Then the next sets on our list would be...
$ \{2\} \to \{2, x_{(2,1)},x_{(2,2)},x_{(2,3)}, \dots\}$
$ \{1,2\} \to \{1,2, x_{(3,1)},x_{(3,2)},x_{(3,3)}, \dots\}$
$ \{3\} \to \{3, x_{(4,1)},x_{(4,2)},x_{(4,3)}, \dots\}$
$ \{1,3\} \to \{1,3, x_{(5,1)},x_{(5,2)},x_{(5,3)}, \dots\}$
$ \{2,3\} \to \{2,3, x_{(6,1)},x_{(6,2)},x_{(6,3)}, \dots\}$
$ \{1,2,3\} \to \{1,2,3, x_{(7,1)},x_{(7,2)},x_{(7,3)}, \dots\}$
$ \{4\} \to \{4, x_{(8,1)},x_{(8,2)},x_{(8,3)}, \dots\}$
And so on forever.
It is then easy to imagine that all the finite subsets of our power set will be on the list, but will there be any infinite subsets that can't be on our list?
According to Cantor, we can diagonalize the list of infinite sets and get a set that can't be on our list.
If $1$ is in the first infinite set, then $1$ will not be in our impossible set.
If $2$ is in the second infinite set, then $2$ will not be in our impossible set.
If $3$ is in the third infinite set, then $3$ will not be in our impossible set.
And so on forever.
As an example, we can say that our impossible diagonal set will be...
$\{3,7,13,16,49,51,\dots\}$
Then, our list of infinite sets will contain a set that begins with the number 3, since our finite set $\{3\}$ is mapped to an infinite set that begins with 3.
Also, our list of infinite sets will contain a set that begins with the numbers 3,7 since our finite set $\{3,7\}$ is mapped to an infinite set that begins with 3,7.
Also, our list of infinite sets will contain a set that begins with the numbers 3,7,13 since our finite set $\{3,7,13\}$ is mapped to an infinite set that begins with 3,7,13.
Also, our list of infinite sets will contain a set that begins with the numbers 3,7,13,16 since our finite set $\{3,7,13,16\}$ is mapped to an infinite set that begins with 3,7,13,16.
And so on forever.
Then, if our impossible set is not in the list, it must contain a number that is greater than any number that appears in the finite sets. And this is impossible.
Is this diagonal set in the list or not?
The cardinality of the infinite subsets may still be greater than the finite subsets, but is diagonalization sufficient to prove it?
There's no basis for this conclusion in your arguments. You've shown that for each $n$, your list contains a set that has the same least $n$ elements as the missing set. That doesn't mean that the missing set is identical with any of these sets, or that it needs to contain any particular numbers in order to keep differing from these sets in each step.
As an example, consider that the set $\mathbb N$ itself may be missing in your list, and convince yourself that none of the sets on your list need be $\mathbb N$, despite the fact that the list contains sets with all finite prefixes of $\mathbb N$.