Suppose that your two reals are very close, so they are equal up to the 100th decimal digit. They differ in the subsequent digits; for example, for one of them the 101th digit could be 3 and, for the other, 7. You could find a rational number in-between, say, the one whose 101th digit is 6 and all subsequent digits are zero.
But, of course, that is not the only rational you can imagine. You could choose the one whose digits from the 101th position are 51234 and then all zeros. In fact, there is an infinitude of rationals in between.
Now, how many irrationals do you think you could find between the same two numbers? How many non-repeatable sequences of digits can you form after the 101th digit? Another whole infinitude of them.
So, if between your two reals there are infinite rationals, as well as infinite irrationals, what is the argument, again, to say that there are just as many rationals as irrationals?
What I try to say boils down to realizing that, if your two reals are a,b, with a<b, there is a bijection between the interval [a,b] and the interval [0,1], namely the map f:[a,b]->[0,1] defined as f(x) = (x-a)/(b-a). As SteveL27 points out, the rationals may be countable but they do not line up in the ordinary lesser-to-greater order, so there is no such thing as two contiguous rationals, let alone two contiguous reals.