Because there is no bijection between the rationals and the reals, wouldn't that imply that there are some irrational reals with no rational between them, allowing this function to be continuous, at least in places we can't actually compute?
So even though there are uncountable reals, for any real number, you can find rational numbers that get arbitrarily close. You can actually see this pretty easily if you think of reals in decimal notation- sqrt(2) is irrational, but
1.4, 1.41, 1.414, 1.4142, 1.41421, 1.414123, ... etc are all rational numbers that get closer and closer to sqrt(2).
So you should think of real numbers as uncountable, but nonetheless surrounded by rationals no matter how close you zoom in.
Take any irrational numbers x and y, and suppose for convenience that x < y.
Let d = y-x i.e. the length of the interval between x and y.
Let b be an integer so large that 1/b is less than d.
Consider fractions of the form a/b. As a increases, a/b will eventually (strictly) exceed the value of x. Suppose we fix a to be the smallest such value, i.e. (a-1)/b < x < a/b.
Adding d to both sides we get:
a/b - 1/b + d < x + d
Using the fact that 1/b < d and x + d = y, we get a/b < y.
Therefore x < a/b < y.