The problem is you can't just expand the range to infinity. You've got a counting function #{i: i is a natural number, i<k} > #{2i: i is a natural number, 2i<k} for some number k

the range of this "counting" function #{x,..} is the natural numbers. What number will it give when your set {x,...} is infinite?

It is true that in any finite interval you'll have close to twice as many integers as even integers (you can show this ratio holds in the limit with L'Hopital's rule) but that doesn't grant us the ability to compare infinities.

In order to compare the "counts" of infinite sets we generalize the notion of counting. With the counting function defined above, two finite sets A, B are the same size if #A=n=#B for some positive integer n. One way of construction this # function would be to take A={a_1, a_2, ..., a_n} and map each a_i to the natural numbers in sequence: a_1 -> 1, a_2 ->2, ..., a_n -> n, and then look at the largest element in {1,...,n}, namely n, and then do the same thing for B.

How about cutting out the middle man and just mapping each element in A to a unique element in B?

This frees us from the problem of comparing sets by calculating n when there is no finite n.

Of course we're back at the bijection definition. It's confusing because for finite sets, cardinality and "size" seem to mean the same thing, but you get something which feels like a "discontinuity at infinity" when you extend the intervals and for every finite interval one set is bigger than the other...

In short, if you show some property holds for any natural number you like, you've shown it for all finite cases, and don't get the infinite case for free.