Mutually exclusive countable subsets of a countable set This is part of a bigger problem I'm trying to prove, but my argument relies of the validity of the following idea. Note that when I say countable, I don't mean finite -- I mean countable infinity. 
Consider the set of natural numbers, $\mathbb{N}$. Now take a countable subset of them, call it $\mathsf{S}_1$. Then take a countable subset of $\mathbb{N}\setminus{\mathsf{S}_1}$ and call it $\mathsf{S}_2$. Then take a countable subset of $\mathbb{N}\setminus\{{\mathsf{S}_1}\cup{\mathsf{S}_2}\}$ and call it $\mathsf{S}_3$. Continue in this manner to construct $\mathsf{S}_i$'s such that ${\mathsf{S}_i}\cap{\mathsf{S}_j}=\emptyset$ for $i≠j$.
My question is this: is it necessarily the case that eventually there will be finitely many $\mathsf{S}_i$'s with $\mathbb{N}\setminus\bigcup_{i}{\mathsf{S}_i}={P}$ where $P$ is either empty or finite? My guess is yes, but I'm not how to prove it or how to find a counterexample.

Edit: @ErickWong is right (see his comment below). This doesn't answer the question. I did cover myself with the caveat "If I understand ...". I may leave this nonanswer up for a while since others may learn from it.
If I understand your question correctly here's another way to show you can have an infinite set left over. Just carry out your construction on the countably infinite set of even numbers. All the odds will remain.
