A direct proof that there exists a real number for which there is no rational number which differs at every point of its decimal expansion

Question

Let $k$ be at least two. There should always exist a real number $r = 0.b_1 b_2 b_3 \cdots$, where this expansion is in base $k$, for which there is no rational number $s = 0.c_1 c_2 c_3 \cdots$ with $b_i \neq c_i$ (again a base $k$ expansion). Why? Well...

Motivation. Suppose one tries to prove that $\mathbb{Q}$ is uncountable, using the same argument of that in $\mathbb{R}.$ So take an arbitrary listing of decimal expansions which are in $\mathbb{Q}$, and consider the diagonal. The diagonal is just an expansion of a real number $r$, and if one could always find a rational number which differs in each place of $r$, then one could prove that $\mathbb{Q}$ is uncountable since one would find a rational number not on the list.

So this shows that there must exist a real number $r$ for which this is not possible.

I would like a direct proof of this fact, for any base $k$. In particular, an example of such an $r$. For example, if one takes $k=2$, the result is trivial: the only real number which differs in every decimal place is simply $1-r$, which is rational iff $r$ is, so taking any irrational $r \in (0, 1)$ suffices.

Answer 1

Enumerate the rational numbers, $\{q_i\}$. Define $r$ such that the $i^{th}$ digit of $r$ equals the $i^{th}$ digit of $q_i$. This number satisfies the requirement.

Answer 2

Almost every real number in $[0,1)$ is normal in base $k$, meaning every possible finite length string of length $n$ occurs in the base $k$ expansion with probability $\frac{1}{k^n}$. So take any normal number $x$ and any rational number $r$. Then $r$ repeats with a certain period after that point. Say the period has length $n$ and contains the digit $d$ (for $0 \le d < k$). Then there is a string $dd\dots d$ of length $k$ appearing in the expansion of $x$, so $r$ and $x$ have a point in common.