This appears to be common doubt expressed in the following places:

What are the cases of not using (countable) induction?

Why induction cannot be used for infinite sets?

Why doesn't induction extend to infinity? (re: Fourier series)

Induction essentially shows if something is true for 1 and n, it is true for n+1 - and is thus true for countably inifinite cases (bijection with N)

Stephen Abbott has a an exercise in Chapter 1 (1.2.12) that suggests that one cannot use induction to prove that a countable union of countable sets is countable. For clarity lets assume countable is countably infinite.

Rudin in chapter 2, statement after (3) clearly states that the notation for Union over the set A of all positive integers uses the infinity symbol. He states that he uses the infinity symbol for a union of sets to indicate a countably infinite union of sets and distinguishes it from the infinity used to extend the reals.

$N$ is countably infinite by definition. Therefore any bijection with $N$ is also proved for countably infinite cases.

Response 1:
One answer is that n=infinity cannot be demonstrated via induction, as infinity is not a natural number - there is no n+1 = infinity. But: The infinity symbol is not used in this way as per Rudin. He is using infinity in place of N. There is a problem in semantics - see response 3.

Response 2: You don't need induction so why do you want to use it here to prove the statement. Well there is as much to learn from why not as from why.

Response 3: Induction proves for finite cases only. It only says we prove for every n in N. Whatever n you pick is by definition finite. So we have only proved finite unions of countable sets are countable. **While induction proves this for countably infinite cases - each case is finite.** In this sense response 1 makes sense - no case is anything but finite. Could the smarties on this thread validate this reasoning?

wow 2 downvotes but no explanation why...?