Doesn't the unprovability of the continuum hypothesis prove the continuum hypothesis?
Here's an example axiomatic system:
- There exist exactly three objects $A, B, C$.
- Each of these objects is either a banana, a strawberry or an orange.
- There exists at least one strawberry.
Let's name the system $X$.
Vincent's Continuum Hypothesis (VCH): Every object is either a banana or a strawberry (i.e., there are no oranges).
Now, to disprove this in $X$, you would have to show that one of $A, B, C$ is an orange ("construct a counterexample"). But this does not follow from $X$, because the following model is consistent with $X$: A and B are bananas, C is a strawberry.
On the other hand, VCH does not follow from $X$ either, because the following model is consistent with $X$: A is a banana, B is a strawberry, C is an orange.
As you can see, there is no contradiction, because you have to take into account different models of the axiomatic system.
I think the basic problem is in your statement that "In order to disprove it, one would only have to construct one counterexample of a set with cardinality between the naturals and the reals." Actually, to disprove CH by this strategy, one would have to produce a counterexample and prove that it actually has cardinality between those of $\mathbb N$ and $\mathbb R$.
So, from the fact that CH can't be disproved in ZFC, you can't infer that there is no counterexample but only that no set can be proved in ZFC to be a counterexample.
How would you prove that not all groups are abelian? Namely, how would you prove that the axioms of groups do not imply $\forall x\forall y(xy=yx)$?
Well. You'd find a model for the theory, namely a group, which is not abelian. At the same time, you can also show this is not disprovable by finding an abelian group.
So, does that mean that all groups are abelian, because the axiom $\forall x\forall y(xy=yx)$ is independent of the axioms of a group? No. It does not.
The independence of the continuum hypothesis from ZFC goes along the same lines. We can show there are models of ZFC where CH holds, and others where it fails. This is regardless to the techniques we use, which are clever and useful for so much more in set theory.