Firstly, you should not say "infinite integers" if you mean "infinitely many integers". If one admits such a thing as an infinite integer, and $n$ is an infinite integer, then $n$ and $n+1$ are infinite integers, but they are not infinitely many, since there are only two of them. This is standard usage in mathematics, regardless of what usages may prevail in informal English used in contexts other than mathematics.
The sequence below has a first term, a second term, a third term, and so on, so there are just as many terms as there are positive integers $1,2,3,\ldots\,{}$.
\begin{align}
\frac 1 2, \underbrace{\frac 1 3, \frac 2 3},\underbrace{\frac 1 4,\frac 3 4},\underbrace{\frac 1 5,\frac25,\frac35,\frac45},\ \underbrace{\ldots\ldots\ldots\ldots}_\text{sixths},\ \underbrace{\ldots\ldots\ldots\ldots}_\text{sevenths},\ \ldots
\end{align}
And all rational numbers between $0$ and $1$ are in this list (you'll notice I skipped $2/4$ since it's not in lowest terms).
If you want to add one more rational number, or two more (e.g. $0$ and $1$) then just append them to the front of the list and you'll see that there are still just as many as there are integers.
This differs from your argument and doesn't directly answer your question, but I hope it clears up a couple of confusions that might get in the way of your thinking about this.
"We start with the fact that between any two real numbers there is a rational number."
True. But not just one; there are an infinite number of rational numbers between two real numbers.
"There is an infinite amount of real numbers between any two real numbers."
True. There are an infinite amount of real numbers including an infinite amount of rational numbers between two real numbers.
" Hence any real interval can accommodate the whole set of rational numbers which is also infinite."
Well, it can contain a set of the same cardinality as the whole set of rational numbers. We'll call that "accomodating". (You can't actually put 1/5 into [37.5, 37.8]. But you can put $\phi(1/5)$ into [37.5,37.8] when $\phi$ is a bijection from Q to $Q \cap [37.5,37.8]$.) It can also accommodate the whole set of real numbers between to real numbers. (That is we can make a bijection, $\rho$ where $\rho(x) \in [37.5,37.8]$ for all $x \in \mathbb R$.)
"We know there is a bijection between the set of natural numbers and the set of rational numbers."
True.
" Hence if there were a "bijection" between the natural and real numbers the domain (the natural numbers) would be mapped entirely within this interval of real numbers to all the rational numbers within."
Yes. And what's wrong with that?
Between x and y there are an infinite number of rationals. There is a bijection, $\phi$, between $Q$ and the rationals between x and y. And There is a bijection, $\chi$ between $N$ and $Q$. So $\phi\circ\chi$ is a bijection between $N$ and the rationals between x and y. Nothing wrong with that.
"So there cannot be any natural number mapping to something outside of this real number interval so there cannot be a bijection so the set of real numbers is uncountable. "
Sure there can. That would be just a different mapping. Having one map map to somewhere doesn't "use up" the ability to have another map mapping elsewhere.
Consider $f(n) = 2n + 1$. This is a 1-1 map from the natural numbers to the odd numbers. There is "nothing left" to map to any even number. Yet $g(n) = n$ maps to all the even numbers including the odd numbers. This isn't a contradiction.
======
There is nothing in this proof that pertains to the real numbers that doesn't pertain to the rationals as well. So whatever this proves about the reals it also proves about the rationals. As the rations are countable this can not prove the reals are countable so there must be an error.
The error seems to be that you assume if you have a bijection $\phi: N \rightarrow S$ then you can not also have a bijection $\rho: N \rightarrow T$ where $S$ is a proper subset of $T$.
That is simply not the case on infinite sets. We have bijections from that Natural numbers to the integers, the rationals, the even numbers, etc. many of which are subset and supersets of each other.
In particular we can and do have bijections between the natural numbers and the rationals in any interval.
Let [x,y] be an interval. And let x< q < r < y so that q and r are rational. Let $f(t) = (t/t+1)(1/2(r - q)) + q + (r-q)/2$. $f$ is just such a injection from all rationals to the rationals in [x,y].
Best Answer
The standard proof that the real numbers are uncountable does not need to be phrased as a proof by contradiction. To say that $\mathbb{R}$ is uncountable is to say that if $f : \mathbb{N} \to \mathbb{R}$ is a map, then it is not a bijection. The standard proof in fact proves that $f$ is not a surjection by explicitly exhibiting a real number which is not in the image of $f$.
As Daniel says in the comments, the problem with your proof is that there's no reason the number you want to write down has to be rational, and that's it. As I've said in several other answers recently, the problem with proofs by contradiction is that once you've made a mistake you think you're done. Generally it's better to prove things directly.