Cantor's Diagonal Argument Why Cantor's Diagonal Argument to prove that real number set is not countable, cannot be applied to natural numbers? Indeed, if we cancel the "0." in the proof, the list contains all natural numbers, and the argument can be applied to this set.
 A: How about this slightly different (but equivalent) form of the proof?  I assume that you already agree that the natural numbers $\mathbb{N}$ are countable, and your question is with the real numbers $\mathbb{R}$.
Theorem: Let $S$ be any countable set of real numbers.  Then there exists a real number $x$ that is not in $S$.
Proof: Cantor's Diagonal argument.  Note that in this version, the proof is no longer by contradiction, you just construct an $x$ not in $S$.
Corollary: The real numbers $\mathbb{R}$ are uncountable.
Proof: The set $\mathbb{R}$ contains every real number as a member by definition.  By the contrapositive  of our Theorem, $\mathbb{R}$ cannot be countable.
Note that this formulation will not work for $\mathbb{N}$ because $\mathbb{N}$ is countable and contains all natural numbers, and thus would be an instant counterexample for the hypothesis of the natural number version of our Theorem.
A: Cantor's diagonal argument is one of contradiction.  You start with the assumption that your set is countable and then show that the assumption isn't consistent with the conclusion you draw from it, where the conclusion is that you produce a number from your set but isn't on your countable list.  Then you show that for any 
If you wanted to do Cantor's diagonal argument, you'd start by assuming the Natural numbers are countable, then try to create a contradiction by producing a number that isn't on the countable list.
Suppose the natural numbers are countable, and we list them then perform Cantor's diagonal technique, producing a string of digits $x$.  $x$ either has an infinite number of non-zero digits or $x$ has a finite number of non-zero digits.
If $x$ has a finite number of non-zero digits, then $x$ is on the list (and it's a Natural number), so there no contradiction.  If $x$ has an infinite number of non-zero digits, then $x$ isn't a Natural number (and it isn't on the list), so there is no contradiction.
Either way, we can't find a contradiction by using Cantor's diagonal argument on the assumption that the Natural numbers are countable, so we can't conclude from that argument that the Natural numbers are uncountable.
The most important part here is to notice that any natural number must have a finite number of non-zero digits.
A: The list contains all natural numbers, but also quite a few more.
Natural numbers are terminating strings of digits, that is they are of finite length. Cantors diagonal argument enumerates all strings of digits, especially non-terminating ones.  And yes, the set of those is uncountable, whereas the set of terminating strings is in indeed countable.
A: Make a list of natural numbers, put into separate rows, as in setup for the Cantor's argument. Applying the diagonal argument extract a new number different from each row. The main, or even maybe the only, reason why this does not disprove countability of natural numbers is that this new constructed number is not necessarily a natural number. True, you have found a new number not among your list, but it is not necessarily a new addition to natural numbers.
Here is an analogy: 
Theorem: The set of sheep is uncountable. 
Proof: Make a list of sheep, possibly countable, then there is a cow that is none of the sheep in your list. So, your list could not possibly have exhausted all the sheep!
What is wrong with this proof? (Unless you believe there are uncountably many sheep! :))
