Taking away infinitely many elements infinitely many times This is a somewhat hand wavy question but I'm not sure how to ask it more precisely.
If we have a countably infinite sequence (or set), can we take away infinitely many elements from the sequence infinitely many times and still be left with infinitely many elements? Or is it the case that we must be left with 0 or finitely many elements?

Here's a construction on the natural numbers.  
First remove all powers of $3$.  Then remove all powers of $5$.  Then remove all powers of $7$.  Then remove all powers of $11$.  Continue in this way, removing all powers of each odd prime.  Clearly these removed sets are disjoint (a number cannot be a power of both $3$ and $5$, for example), and infinite.  
But what we have left is also infinite.  For example, it contains all powers of $2$, and this is an infinite set.

Another possibility (not using any facts from number theory) - let $S$ be the set of all finite $0-1$ sequences (for example, $010001110, 1001, 000\in S$).  Remove all sequences starting with $01$.  Then remove all sequences starting with $001$.  Then remove all sequences starting with $0001$, and continue in this way.  Once again, we remove infinitely many elements infinitely many times, but what we have left (namely, the set of sequences starting with a $1$, plus the constant $0$ sequences) is still infinite.

Update: You have assumed in your question that it is possible to remove infinitely many elements from a set infinitely many times.  Well, if you can do that (and you can), then just leave off removing the first infinite set.  Then what you have left at the end will be infinite.
