In your set of positive even numbers example, you are only showing that a specific enumeration scheme does not generate the set of even numbers. A diagonalization argument requires that you show that no such enumeration scheme can exist (not just a particular chosen one). In the typical proof of diagonalization, it does not presuppose any particular scheme of listing the numbers. It just assumes that some arbitrary one exists, then derives a contradiction.

For your binary tree construction, a trancedental real number can be represented by a path of infinite length down the tree. Your argument is that since a breadth first traversal will eventualy exhaust that whole path, that the real number described by it will have been encountered. There are two ways to interpret what you are doing wrong:

- If we are indexing/pairing these nodes by time steps (an index), your construction is using a countably infinite time step to express the numbers described by an entire path (which defeats the point of being countable).

- For countable sets, you have to give me an index of finite size. If I give you a real number, you need to return a natural number (or equivalent) that indicates where it is. To test this, give me the index of pi in your claimed "countable" enumeration. The reason you wont be able to somewhat follows from Cantor's diagonalization scheme.

You might want to do some Googling before spending the time to write up an entire blog post about it (god forbid posting it to HN). The mistake you made is very common and has been discussed to death. You would have caught it.