Are there uncountably many cube-free infinite binary words?

There are uncountably many cube-free infinite words. Indeed, consider any cube-free infinite word in 2 letters $a$ and $b$ (say, the Thue-Morse word). This word contains infinitely many occurrences of $a $. Replace some occurrences of $a$ by $a'$ and some occurrences of $a$ by $a''$. You get a new infinite word in $a',a'',b$ which is also cube-free (but in 3 letters), a continuum of them. One can then use a substitution from a 3-letter alphabet to a 2-letter alphabet that preserves cube-freeness (see Bean, Dwight R.; Ehrenfeucht, Andrzej; McNulty, George F.Avoidable patterns in strings of symbols. Pacific J. Math. 85 (1979), no. 2, 261–294) to obtain a continuum of cube-free words in 2 letters.

Denote by $\mu$ the mapping from the Thue-Morse sequence, $\mu(0)=01$ and $\mu(1)=10$. Now define a sequence of maps from binary words to binary words, $g$, so that $g_{\emptyset}(w)=w$, $g_{0b}(w)=\mu^2(g_{b}(w))$ and $g_{1b}(w)=0\mu^2(g_{b}(w))$. Now given an infinite binary sequence $B=b_1b_2\dots$ define $w_{B}$ to be the limit of $$g_{b_1}(w),g_{b_1b_2}(w),g_{b_1b_2b_3}(w),\dots$$ The $w_B$ give you uncountably many $7/3$-power free words (so in particular, cube free) which moreover have infinitely many overlaps.

This stronger result is proved here. I believe all known constructions of large families of such sequences are defined by iterative mappings.

One can also consider the following. Let x be the Thue Morse sequence. Let X be the closure of the set of shifts of sequences i.e. sequences obtained by deleting the first few letters. This set is perfect. Hence X is uncountable. Also every element of X is cube free.
The only thing to check here is that X is perfect. For this it is sufficient to check that x is a limit point. One can do this by using the fact that x is generated by the sequence $0 \rightarrow 01$ and $1 \rightarrow 10$. The topology on $\{0,1\}^{\mathbb N}$ is the product of the discrete topology on $\{0,1\}$ .