Cohen's result is that from a certain set of axioms ($\mathsf{ZFC}$) we cannot prove the continuum hypothesis. Gödel's result is that from the same set of axioms, we cannot refute the continuum hypothesis. This only means that the set of axioms under consideration is not strong enough to settle this question. If you manage to exhibit a set of intermediate size, your argument necessarily uses axioms beyond those in $\mathsf{ZFC}$, or is not formalizable in first-order logic. Similarly if you manage to show that there is no such set. It is worth pointing out that there are standard axioms beyond those in $\mathsf{ZFC}$ that settle the continuum problem. These axioms are not universally accepted yet, but this illustrates that, as explained, their results are not absolute, but relative to a very specific background theory.
Throughout this answer I'm assuming ZFC is consistent.
You're essentially asking what the interval between $\aleph_0$ and $2^{\aleph_0}$ can be. E.g. can there be two "intermediate cardinalities"?
The answer is basically: the continuum can be anything whatsoever, with two exceptions: obviously it has to be uncountable, and it turns out it also has to have uncountable cofinality.
For example, it's consistent with ZFC that $2^{\aleph_0}=\aleph_{17}$, or that $2^{\aleph_0}=\aleph_{\omega^2+42}$. It's even consistent with ZFC that $2^{\aleph_0}$ is a "fixed point of the $\aleph$-function," that is, it's consistent with ZFC that $2^{\aleph_0}=\aleph_{2^{\aleph_0}}$. (Fine, more carefully: it's consistent with ZFC that $2^{\aleph_0}=\aleph_\lambda$ where $\lambda$ is the initial ordinal of $2^{\aleph_0}$.)
This isn't very precise, of course, and has some obvious issues - e.g. we clearly can't have $2^{\aleph_0}$ be the smallest cardinal of uncountable cofinality bigger than the continuum, even though the latter is obviously a cardinal of uncountable cofinality! Unfortunately, it takes real work to make the above precise: we need to talk about models of set theory, and in particular forcing:
Suppose $M$ is a model of ZFC + CH and $\kappa$ is a cardinal of uncountable cofinality in the sense of $M$ (or more generally, $M$ is any model of ZFC and $\kappa$ is a cardinal of uncountable cofinality in the sense of $M$ which is bigger than $M$'s continuum). Then there is a forcing extension $N$ of $M$ with the same cardinals and cofinalities satisfying $2^{\aleph_0}=\kappa$.
The requirement that $\kappa$ have uncountable cofinality is necessary as a consequence of Konig's theorem: ZFC proves that the continuum has uncountable cofinality. The above result is due to Solovay, and has a terrific strengthening to the powersets of all regular cardinals due to Easton. Singular cardinals, meanwhile, turn out to be wildly more complicated.
Let me make a couple footnotes about forcing:
First, note that I'm playing fast and loose with the assumption that forcing extensions exist, above: we either need to assume that $M$ is countable, or use the Boolean models approach to forcing. But that's a secondary issue.*
Second, remember that forcing is definable: the bolded fact above can be turned into a ZFC theorem saying roughly "For any cardinal $\kappa$ of uncountable cofinality, there is a cardinality- and cofinality-preserving forcing making $\kappa$ the continuum." (And in fact we know what forcing does the job.) While talking about models makes everything much easier to understand, there is an actual "internal theorem" here.
Best Answer
On one hand, sure, there are such sets. For example, you can construct one by following the proof of Hartogs's theorem. Hartogs's theorem gives that the set of all order types of well-orderings of subsets of $\mathbb{N}$ has cardinality $\aleph_1$ exactly (and is therefore uncountable). However, there are only continuum many relations on subsets of $\mathbb{N}$, so we can code a representative of each of these $\aleph_1$-many order types as a real number. Here, a representative just means some relation with that order type. This gets you a subset $\mathcal{W}$ of $\mathbb{R}$ with cardinality exactly $\aleph_1$. The coding can be done fairly explicitly (although I will leave that as an exercise in this answer).
Since the Continuum Hypothesis is independent of ZFC, it's clear that ZFC cannot prove that $\mathcal{W}$ (a set with cardinality $\aleph_1$) has the same cardinality as $\mathbb{R}$ (which has cardinality $2^{\aleph_0}$).
On the other hand, your explicit question was whether it follows from [the fact that the Continuum Hypothesis (CH) is independent of ZFC and therefore there exist models of ZFC in which the CH is false] that there must exist an uncountable subset of $\mathbb{R}$ which it is impossible to prove has the same cardinality as $\mathbb{R}$. And the answer to this question is a bit more delicate. In the colloquial sense, this certainly does not follow merely from the fact that CH is independent of ZFC.
But practically, much easier, slightly dodgy answers could have worked: consider e.g. the set defined using Separation as $\mathbb{N} \cup \{x \in \mathbb{R} \:|\: x = x \text{ and CH holds}\}$, which has the same cardinality as $\mathbb{R}$ precisely if CH holds, and smaller cardinality otherwise - but then not cardinality $\aleph_1$ (instead, you can prove it countable then). If you want to understand in precisely what way such answers are dodgy, you have to grapple with some borderline philosophical questions, such as what it means to give an example of a set. But that's a story for another time.