TITLE: Mutually exclusive countable subsets of a countable set
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: It's easy to construct counterexamples in a variety of ways, as the other answers show. I wanted to point out that once you construct some countably infinite $S_1$ with $\mathbb{N}\,\backslash\,S_1$ also countably infinite, then you can construct each subsequent $S_i$ in the same manner that you constructed $S_1$.
For example, suppose you start with $\mathbb{N}$, and choose $S_1=\{0,2,4,6,8,\ldots\}$, i.e. the even numbers.
Then you can take the set $\mathbb{N}\,\backslash\,S_1=\{1,3,5,7,9,\ldots\}$, order it in an increasing sequence, and then let $S_2$ be the elements of that sequence there with an even index, i.e. $S_2=\{1,5,9,13,\ldots\}$. Then you can repeat the process to construct $S_3,S_4,\ldots$ ad infinitum.

I'll express this idea more formally:
Since $S_1$ has the same cardinality as $\mathbb{N}$, there is some bijective function $f:\mathbb{N}\to S_1$, for example the function $f(n)=2n$.
Since $\mathbb{N}\,\backslash\,S_1=\{1,3,5,7,\ldots\}$ also has the same cardinality as $\mathbb{N}$, there is some bijective function $g_1:\mathbb{N}\to\mathbb{N}\,\backslash\,S_1$, such as $g_1(n)=2n+1$.
Now we can construct
$$S_2=\{g_1(f(n))\,|\,n\in\mathbb{N}\}=\{1,5,9,13,\ldots\}$$
And you can continue to construct $S_3$, given a function $g_2:\mathbb{N}\to\mathbb{N}\,\backslash\,(S_1\,\cup\,S_2)$, such as $g_2(n)=4n+3$:
$$S_3=\{g_2(f(n))\,|\,n\in\mathbb{N}\}=\{3,11,19,27,\ldots\}$$
And in general for any $k$ you can construct $S_{k+1}$, given a function $g_k:\mathbb{N}\to\mathbb{N}\,\backslash\,(S_1\,\cup\,\ldots\,\cup\,S_k)$:
$$S_k=\{g_k(f(n))\,|\,n\in\mathbb{N}\}$$