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?

WLOG assume X in [0,1]

Pick a sequence of I_n s/t I_n is either 1 or 0 for all n = 1, 2, 3...

Let your A_n be the sum from 1 to n of I_n / 2^n

It should be totally obvious that
1) A_n is monotone nondecreasing and rational
2) You can always pick I_{n+1} s/t A_{n+1} <= X
3) If A_n =/= X you can find N>n s/t A_n + 1/2^N is still <= X
4) 1-3 imply that you can always find an n' such that X - A_n' < E for all E in R (if A_n doesn't if your bill, find additional N until it does)

Voila.  A sequence of rationals that converges to any X in [0,1] you want.  For X not in [0,1] add the integer part of X, which is rational to the front of your sequence, and take negatives if you need to.

Really I think all you need to do is construct the monotone sequence, bound it by X, and invoke monotone convergence.