A proposition...

For all x in R, there exists a sequence a_n in Q such that the limit of a_n as n->+inf = x

This seems trivially false to me just from cardinality arguments (e.g. if you assume it's true then the cardinality of Q being less than the cardinality of R immediately implies that there are two different x in R that must have the same limit in Q), but I'll be damned if I can prove it rigorously.  Is there an easier way?

>>14
A countable product of countable sets is not necessarily countable, as exemplified by the very case we are discussing. The proof that a countable union of countable sets is countable is a trivial extension of the fact that N x N is countable.

In general, {sequences of elements of S | cardinality of S > 1} is an uncountable set, by a diagonalization proof:
Suppose it's countable, enumerate the sequences:
x11, x12, x13, ...
x21, x22, x23, ...
x31, x32, x33, ...

Create a sequence y1, y2, y3, ... in which yi != xii. This is obviously possible since each element of the sequence can be any element of S, and there are at least two elements of S.
Clearly y1, y2, y3, ... is not an element of the supposed enumeration of sequences, and therefore our assumption that the set of sequences is countable is invalid.