I have spent some time (maybe too much) thinking about the CH.
Intuitively, I would say that there is nothing between $\aleph_{0}$ and ${\mathfrak c}$, just as there is no Integer between 0 and 1. - No proof, of course.
So I thought about fractal subsets, based on the idea: "If fractals can have non-integer dimensions, maybe they can help with this too".
Of course, for the Cantor set, there is an easy proof that it's uncountable, so this doesn't help either. Same thing applies for similar sets. Still, I thought that a fractal approach might help with this question.
But then I read up about constructibility and found (paraphrased): Creating fractal subsets of $\mathbb{R}$ would always result in a constructible set, which doesn't prove anything, since we have no proof that all sets are constructible.
Is this correct?
What you're saying is not quite correct.
Fractal, for the most case (that you define them), are just going to be Borel or something close enough. Borel sets are not constructible sets. They are, however, have the perfect set property. Therefore Borel sets cannot be a counterexample to CH in any case.
It is possible, though, that the set of the constructible reals (which itself is a constructible set) is a counterexample to the continuum hypothesis. For example, start with $V=L$, and then add $\omega_2$ Cohen reals.
What you do say, however, is somewhat correct, sets which are produced from a "naive definition over the reals" are unlikely to be capable of serving as a counterexample of the continuum hypothesis.