Cantor’s diagonal proof revisited In his diagonal argument (although I believe he originally presented another proof to the same end) Cantor allows himself to manipulate the number he is checking for (as opposed to check for a fixed number such as $\pi$), and I wonder if that involves some meta-mathematical issues. 
Let me similarly check whether a number I define is among the natural numbers. The number is $n+1$ and it is clear that $1$, $2$, $3,\ldots,\ n$ are not among these numbers. This “proves” that $n+1$ is not a natural number.
I have here, just like Cantor, a formula for a number, rather than a given number. What is the difference between our proofs?
It seems to me that the answer is that Cantor’s number (as opposed to mine) is being successively better bounded; the process of moving forward in his enumeration describes a converging series, such as we use to define the real numbers, using Cauchy limits. You don’t hear this added comment in the proof. Don’t you think it belongs to the proof? 


Let me similarly check whether a number I define is among the natural numbers. The number is n+1 and it is clear that 1,2,3 …n are not among these numbers. This “proofs” that n+1 is not a natural number.

No, all you have proven is that there is a positive integer ($n+1$) that is not in the list  $1, 2, \ldots, n$.
Such a fact does make for a perfectly good proof that $1, 2, \ldots, 73$ is not a list of all of the positive integers. It is even a proof that no list of the form $1, 2, \ldots, n$ (where $n$ is a positive integer) can be a list of all of the positive integers.
If the set of positive integers were finite, it would be of the form $1, 2, \ldots, n$. So it can even be used as a proof that the set of all positive integers is infinite.

All of that is analogous to the proof you reference. The diagonal argument (constructively) proves that every list of real numbers is missing a real number.
Thus, we conclude that there does not exist a list of all real numbers.
