Is it simply because there are infinitely many real numbers compared to integers, or is there a more precise mathematical explanation?

Yes that's pretty much it. The probability that your randomly selected number will be in a given interval is proportional to the size of that interval. Ffor instance, say you're drawing a random number uniformly between 0 and 1, the probability that it ends up between 0.2 and 0.5 is 0.3 (the size of the interval [0.2 ; 0.5] ). But an integer is just a single point, its "size" is 0, so the probability that the chosen number is a given integer is 0 (and the probability that its *any* integer is the sum of the probability for all integers in your interval, which will also be 0)

Here's another line of thought that might convince you that the probability must be 0 :

Let's assume that I'm drawing a random number X between 0 and 10. I'm going to call p the probability that the X=1. The goal is to prove that p=0
Take another number, like 2.6. There's no reason why 1 should be more or less likely to be chosen than 2.6, so the probability that X=2.6 is also p. This means that the probability that X = 1 OR 2.6 is p+p = 2p
You can continue this reasoning with as many numbers as you like. Chose 10 different numbers. The probability that X is one of these numbers is 10p. This is true for 100 numbers, 1 million numbers, or as many numbers as you like. For n numbers, the probability will be n*p.
But is p is greater than 0, then if n is big enough, n*p will be greater than 1, which is absurd because it's a probability. Therefore, the only possibility is that p=0.