We can take a small twist of Matt N.'s hint to make sure we don't run into trouble with dual representations. To see what the problem is, note that, for example, there are two ways of writing $\frac{1}{2}$ in binary:
$$0.100000\ldots\quad\text{and}\quad 0.011111\ldots$$
In principle, it would be that your list begins with $0.1000\ldots$, and then the numbers in position $n$ all have $n$th digit equal to $0$ for all $n\gt 1$. Then the straight "change the digit" process leads to $0.01111\ldots$, which is on the list (it's a different representation of the first number on the list).
So the first thing we need to do is specify that we will use one particular representation in our "list"; usually the one that terminates if there is a choice.
That done, instead of looking at the single diagonal digit, work with diagonals in blocks of $2$; that is, look at the digits in position $1$ and $2$ first, then the digits in positions $3$ and $4$ for the second number, and so on, looking at the digits in position $2k+1$ and $2k+2$ for the $k$th number. Then make the "switch" ensuring that the number you construct does not have two different representations. For instance, if the $k$th number in the list has $00$, $01$, or $11$ in the $2k+1$ and $2k+2$ positions, then put $10$ in your number on those positions; if the $k$th number in the list has $10$, then put $01$ on your list. Then the number you construct does not have two representations, so you can test that it is not on the list by simply comparing the $2k+1$ and $2k+2$ digits with the $k$th number on the list.
So if your list looks like:
$$\begin{align*}
&0.{\color{red}{a_{11}a_{12}}}a_{13}a_{14}a_{15}a_{16}\cdots a_{1,2k+1}a_{1,2k+2}\cdots\\
&0.a_{21}a_{22}{\color{red}{a_{23}a_{24}}}a_{25}a_{26}\cdots a_{2,2k+1}a_{2,2k+2}\cdots\\
&0.a_{31}a_{32}a_{33}a_{34}{\color{red}{a_{35}a_{36}}}\cdots a_{3,2k+1}a_{3,2k+2}\cdots\\
&\vdots\\
&0.a_{k1}a_{k2}a_{k3}a_{k4}a_{k5}a_{k6}\cdots {\color{red}{a_{k,2k+1}a_{k,2k+2}}}\cdots\\
&\vdots
\end{align*}$$
You then take each red block and construct a new number
$$0.\color{blue}{b_{1}b_{2}}\color{green}{b_3b_4}\color{brown}{b_5b_6}\cdots\color{magenta}{b_{2k+1}b_{2k+2}}$$
where $\color{blue}{b_1b_2}$ is chosen so that it is different from $\color{red}{a_{11}a_{12}}$; $\color{green}{b_3b_4}$ is chosen so that it is different from $\color{red}{a_{23}a_{24}}$; $\color{brown}{b_5b_6}$ is chosen so that it is different from $\color{red}{a_{35}a_{36}};\ldots,\color{magenta}{b_{2k+1}b_{2k+2}}$ is chosen so that it is different from $\color{red}{a_{k,2k+1}a_{2k+2}}$, etc. So you are "going down the diagonal" and changing things so that the resulting number is not on the list: it's not the first number on the list, because it differs from it in the first two entries and your number has only one representation. It's not the second number on the list because it differs from that one in the third and fourth digits; given any $k$, the number constructed is not the $k$th number on the list because it differs from it in the $2k+1$ and $2k+2$ positions. Etc.
In any base $b$, using $h$ to denote the highest possible digit, all the numbers in $(0,1)$ that have more than one representation have the form
$$0.xxx\dots p000\ldots=0.xxx\dots qhhh\dots$$
where $p\ne0$ and $q=p-1$. In other words, they have a terminating expansion.
Thus the set of all such numbers is countably infinite by the following bijection between the terminating expansions and the positive integers:
$$(0.x_1x_2\dots x_k)_b\to(x_k\dots x_2x_1)_b$$
(C) is correct for both questions.
Best Answer
If there are only finitely many zeros in each expansion, then each number in the list is a rational number. Hence the list will be countably infinite.