Does Cantor's Diagonal argument prove that there uncountable p-adic integers?

Question

Can I use the argument for why there are a countable number of integers but an uncountable number of real numbers between zero and one to prove that there are an uncountable number of p-adic integers?

My understanding of the proof is that there are an uncountable number of real numbers.

Sequential Integers	Real Numbers*
1	0.346523...
2	0.123456...
3	0.005864...
4	0.089457...
5	0.124598...
6	0.724598...
...	...

*Where none of the decimal expansions terminate in infinitely many 9s

To find an uncounted real number you only need to change one digit in each row to a digit other than a zero or a nine. i.e. 0.436511...

Using the same argument could you not state

Sequential Integers	3-adic Integers
1	...012012
2	...222222
3	...211211
4	...111212
5	...121212
6	...112122
...	...

To find an uncounted p-adic integer you only need to change one digit in each row. i.e. ...202000

Therefore there are an uncountable number of p-adic integers.

Answer 1

As discussed in comments: Yes, Cantor's diagonal argument works exactly the same to show that the set of $p$-adic integers $\mathbb Z_p$ (and hence every set that contains it) is uncountable. If anything, the application in this case is easier than for real numbers, because the series expansions of $p$-adic numbers are unique, whereas in the case of real numbers one has to be a bit careful to deal with the expansions ending in a period of $9999...$.

In fact, the diagonal argument would also show that e.g. the set of power series $k[[X]] = \{ a_0 + a_1 X +a_2 X_2 + ... : a_i \in k \}$ over any field $k$ is uncountable.

Or any set that you can identify with an infinite product $S \times S \times S ...$ of any set $S$ of size $|S| \ge 2$ with itself.