This statement is sometimes used to illustrate the fact that the rational numbers are countable while reals are not. It is obviously true. Considering that obviously true statements in probability theory have a nasty habit of being false there is a point in proving it.

Proposition:
If X is a continuous random variable then P(X ∈ Q) = 0

Proof:
Consider P(X = a). Take e > 0. Since X is a continuous random variable the function F(x) = P(X < x) is continuous. Therefore there exist d > 0 such that F(a+d) - F(a-d) < e. Thus P(X = a) < P(|X - a| < d) < e. Since e is an arbitrary positive number this implies P(X = a) = 0.
Let qn, n ∈ N be a listing of all rational numbers (there are such listings since Q is countable. Then

P(X ∈ Q) = P(UNION(n ∈ N)(X = qn)) = SUM(n ∈ N)(P(X = qn)) = 0 

It is worth noting that the proof relies on the fact that the rationals are countable since a sum over an uncountable set would be meaningless.
A more direct approach would be to write P(X ∈ Q) = ∫ I[x ∈ Q]p(x)dx), where the integral is taken from -∞ to ∞. Unfortunately this is the classical example of a non-integrable function, so unless we opt for some non-Riemann definition of the integral this approach fails.