My try:
Case I
Let $(a,b)$ such that for any subinterval of $C = \bigcap_{m=0}^{\infty}F_m$, $[\frac{n}{3m},\frac{n+1}{3m}]<(a,b)<[\frac{n+2}{3m},\frac{n+3}{3m}]$ for some $n \in \mathbb{N}$
then $C\bigcap(a,b)=\emptyset$
Case II
If $C \bigcap (a,b)\neq \emptyset$ then because we know that for any open interval $(a,b)$ that intersects $C$ contains some subinterval of some $F_m$, the subinterval $[\frac{n}{3m},\frac{n+1}{3m}]⊆(a,b)$.
Suppose that two binary functions in base three $f,g \in[\frac{n}{3m},\frac{n+1}{3m}]$ such that $f(p)=h(p) \; \; \; \forall n≤p≤1$ exists. And because we know that the set of binary sequences having the same initial $n$-segment is uncountable, $C \bigcap (a,b)$ must be uncountable.
I don't know if my analysis is correct (especially Case II). Any hints would be great!
Consider the two halves of Cantor set $C^1_1=[0,1/2]\cap C$ and $C_2^1=C\cap [1/2,1]$. Each of these is uncountable, and they "look the same" (they're homeomorphic, but that's not important) as $C$. So, we can repeat the process and split each half into halves. In general, for each $n\in\mathbb{N}$ you can find uncountable sets $C_i^n$ for $i=1,2,\dots,2^n$ such that each $C_i^n$ has diameter $\le 2^{-n}$ and $C=\bigcup_{i=1}^n C_i^n$.
Granted this, let $I$ be an open interval with $I\cap C\neq \emptyset$. Pick a point $x\in I\cap C$. Let $\varepsilon>0$ be such that $(x-\varepsilon,x+\varepsilon)\subset I$. Pick $n$ with $2^{-n}<\varepsilon$. There exists an $i\le 2^{n}$ with $x\in C^n_i$. Then we have that $C^n_i\subset I$, so that $C\cap I\supset C^n_i$, which is uncountable (in fact of size continuum, but eh).