If irrational numbers are uncountable, then why did I find this? I understand that irrational numbers are uncountable.  I've seen the proof and it makes perfect sense.  However, I came up with this (most likely false) proof that says that they're countable.  Chances are, I made a mistake and the proof doesn't mean anything, but I just want to be sure.  Here is the proof:

You can't find a solid chunk (range of numbers) that does not contain a rational number.  Which means that irrational numbers are just points on the number line, not lines.  Which means you just need to name all the points to count the irrationals.

What is the mistake here?
In mathematical language, the claim that you can't find a solid chunk of the line (meaning an interval) that does not contain a rational is correct, and it is usually stated that the rationals are dense in the line.  

I think your intuition is then that there is one irrational between "neighboring" rationals, which is the sense of saying that the irrationals are points on the line instead of lines.  This is the problem.  Between any two rationals there are an uncountable infinite number of irrationals and a countable infinite number of rationals. When you think of dense sets you need to banish the thought of "neighboring elements" from your mind.  They do not exist because between any two there is another.  

The difference between the rationals and the reals as a field is that the reals are complete.  If you think of the interval $[0,1]$ you can make a binary expansion of a number.  The expansion will converge, but without the completeness axiom you don't know that it will converge to a number in your field.  A classic example is $\sqrt 2 -1$.  If you make its binary expansion the approximation by increasing numbers of bits is a Cauchy sequence.  The reals have an axiom that says it converges to a real, while the rationals do not.  Now the classic Cantor proof applies to say the reals are not countable.
Q: "What is the mistake here?"
A: I believe the mistake is assuming that there is a countable number of segments.

In your 'proof', you say each segment (which contained only rational numbers) is terminated at an irrational number.  Since there are an uncountable number of segments, there are an uncountable number of irrational numbers (each one being at the end of its 'companion' segment).

[EDIT: Removed trailing garbage from earlier edit]