I'm pretty sure that when we say irrational numbers don't repeat, we mean they can't be written like 1/3 = 0.(3), not that a given string of digits can't be found more than once -- if the decimal part of pi is an infinite sequence of digits then surely every substring of digits necessarily repeats infinite times.
But if you mean that the string of digits that make up the decimal part of pi might be an infinitely repeating string, like say if it were 3.(1415), then it wouldn't be irrational because it would be expressible as a ratio. And we know it is irrational: http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational
if the decimal part of pi is an infinite sequence of digits then surely every substring of digits necessarily repeats infinite times
I am not sure at all that this is true. Wikipedia is silent on the issue, so it definitely hasn't been proved (if such a statement gets proved, it will likely make mainstream media world news). I would also speculate your statement has not been disproved either.
There are irrational numbers for which not all sequences of digits appear in their decimal expansion. In fact, one such number was already mentioned in one of the previous posts:
(this irrational number doesn't contain the integers 2,3,..,9 in its decimal expansion at all).
[Edit:] To answer to the OP's original post: no, your idea does not prove that the real numbers are countable. To answer to the OP's original post: +1 to Alrededor's answer.