You aren't undestanding the proof well; the proof isn't flawed. The best way to see it is to read it on a math book with full symbols and so on. Then you can see that there is no flaw, as every step follows from the previous one and that there is no philosophical reasoning about it, only maths.

I think the problem with how you are understanding it is that you think that your enumeration is finite at any time. That's not true.

Let's suppose we are only wanting to enumerate numbers between 0 and 1. So, you have to enumerate them like this:

0 -> some real

1 -> some real

...

n -> some real

for all naturals, so that you have infinite reals in your enumeration scheme. Now, you build your new real number, let's call it R. R is such that its Nth digit after the point is different than the Nth digit after the point of the Nth number. Let's suppose our procedure goes like this:

0 -> 0.1

1 -> 0.11111

2 -> 0.0001

3 -> 0.10101010101010....

4 -> 0.111011

What is R? R is a real number. It has a well-defined value for every one of its digits, so why wouldn't it be a real number? So we have, a real number between 0 and 1, and an infinite enumeration of reals.

Let's see the value of R:

0 -> 0.1

1 -> 0.11111

2 -> 0.0001

3 -> 0.10101010101010....

4 -> 0.111011

so R is:

0.00110 and so on.

Is R 0.00110? R has more digits, is more like 0.00110abcdefg with abcdefg being some more digits we should calculate. R has infinite digits, not only those, and we could calculate all of them by a finite procedure, so we aren't doing anything magical here. What those digits' value is we don't mind, as you'll see.

What we mind about is this: R is different to the first number. How can we know? Because by construction, the first digit of R is different from the first digit of the first number. Now, let's see if R is in our countable set. If R was in our countable set, R would be in a place, for example in the position M of the enumeration. Now, in the position M of the enumeration we have the Mth number, and by construction, the Mth number is different to the R number, as they have at least a digit that is different between them (and we can even say: the Mth digit is different between them).

So, how could R be in the enumeration scheme if we can't place it at any M? Now, that's why we have an absurd here, and that's why every enumeration scheme for reals will be incomplete.

I wish this solves your doubts about it. Please only focus on the mathematical demonstration, we aren't saying nothing about countability, infinities, or very large numbers excepting when we have to.