I came up with a construction of an uncountable set of open balls in $[0,1]\subseteq \mathbb{R}$.
I realize this is wrong (since every open ball contains some $x\in\mathbb{Q}$), but I'd like to know why my argument breaks down? (I will not accept answers which do not address my question)
In a construction of the Cantor Set, we iteratively remove open intervals. In ternary numeral system:
\begin{align} C_1&:=[0,1]\setminus (0.1,0.2)\\ C_2&:=C_1\setminus (0.01,0.02)\setminus (0.21,0.22)\\ \vdots\\ C&=\lim_{n\to\infty}C_n \end{align}
What if at each stage we split the removed intervals in two and took two open intervals. Then for each element $c\in C$, there is some open interval. E.g. For $0.1\in C$, let $(0.1,0.\overline{1})$ be the open interval, or, $0.022\in C$ gives $(0.2\overline{1},0.22)$ is an open interval.
And the Cantor set is uncountable, so there are an uncountable number of intervals?
With your description, $C_1$ has $1$ removed interval, $C_2$ has $2$, $C_3$ has $4$ removed, and so on, in general $C_k$ has $2^{k-1}$ removed intervals. If at each stage you split the removed intervals each into two parts, then what this does is to double the number of removed intervals at each stage, so that at stage $k$ you now have $2^k$ removed intervals.
Note that for this, as in the usual construction of the cantor set, there are only a countable number of removed intervals. So if in some way one associates an element of the Cantor set with each removed interval, one has only associated countably many elements of the Cantor set with the removed intervals. However since the Cantor set is uncountable, there remain (uncountably many) elements of the Cantor set which have not been associated with any of the removed intervals. In short there isn't a contradiction.