With the rationals in a grid, even ignoring that I can trivially arithmetically compute the corresponding natural (interleave the (finite!) digits of the numerator and denominator), I can give a bound on a similar search (say Z-order): O(n²).
I think the fundamental idea you are missing is that a natural number must have a finite number of digits. (This stems from the construction of naturals as finite objects.) Reals are not subject to this limitation, as evidenced by the existence of the irrationals.
[Actually, that last paragraph begs another question: what are the theoretical implications of the fact that any specific real must have a finite representation of some form? Does the reason we can reason about reals which we can't specifically identify with a finite construction have anything to do with the axiom of choice?]
So in my example above, π would not work if, say, your tree was a parse tree of mathematical formulæ: one could locate the formula for π in finite time. Unfortunately I cannot give an example of a single number which does not have such a finite parse tree, as, by its very nature, I cannot write down a finite description of it!
However, if you fix a finite alphabet beforehand, you will never be able to describe all real numbers using strings constructed from this alphabet. You can always add a new symbol to your alphabet to represent some new irrational number, but no matter how many symbols you add you'll never be able to reach "every" irrational number.
Put another way, for a given alphabet there's no way to represent every irrational number in a finite way. However, you could have an irrational number that can be finitely represented in one alphabet, but unable to be finitely represented in another.