real analysis - is it possible to proof that this number is not rational

It is an idea I had when reading the proof that $(0,1)$ is uncountable. There the numbers in $(0,1)$ are written into a list in decimal expansion and then the diagonal is modified and the resulting number is a number not on the list.

Now instead consider $S= \mathbb Q \cap (0,1)$. Like in the proof about $(0,1)$ write the numbers into a list in decimal expansion. Add $1$ to every digit on the diagonal and compute the remainder modulo $10$. I am trying to proof that this new diagonal number is not rational but without using the knowledge that the rationals are countable. Here is the proof:

There are two cases: A rational has either a finite expansion or is periodic. Let the digits in the expansions be called $a_{mn}$. Let the new diagonal number after the modification be $d_n$.

In the first case: If $a_{nn}$ is finite then $d_n$ is finite and $d_n = 0$ for $n>N$ for an $N$. Then, it is possible to find $n+1$ different periodic rational with no $0$ in the expansion. But this is a contradiction. Therefore $a_{n n}$ can not be finite.

In the second case: if $a_{nn}$ is a periodic rational with period $c_1 c_2 \dots c_N$. Then because of a same argument like the finite case this period can not be constant ($N=1$). It is forced to contain all digits in $\{0, ..., 9\}$. But what now? Is it possible to finish this proof without using that the rationals are countable?

Answer

You start by explicitly assuming that $S=\mathbb Q\cap (0,1)$ is countable as you write all these numbers into a list.
If your list does not contain all elements of $S$ it is well possible that the antidiagonal number is in fact one of the rationals not in the list.
For example if $S$ consists only of those rational numbers having at least one $2$ in their decimal expansion, it might happen that the diagonal number is simply $0.222\ldots =\frac29$ and your antidiagonal number becomes $0.333\ldots=\frac13$, which is rational but not an element of $S$.
On the other hand, if $S$ really contains all eventually periodic decimal expansions then it is clear that the antidiagonal is not eventually periodic as it differs from each single element of $S$.

By the way, you should have a closer look at how you define your antidiagonal number: You might accidentally end up with $0.000\ldots=0$ or $0.999\ldots = 1$