The continuum hypothesis can be proved. And disproved, too. Wait, what?
Well, in order to make sense of that, and of the fact that the continuum hypothesis can/cannot be proved/disproved, we first need to understand that proofs don't exist in vacuum. Proofs are sequences of formal statements which include either axioms, or statements derived from previous statements in the sequence. We say that a certain sequence is a proof of a statement, if that statement is the last one in the sequence.
So first, before talking about proving or disproving the continuum hypothesis (or anything else) we need to talk about the axioms, and the inference rules. Well, the inference rules part is somewhat more standard through the most of mathematics, so I will kindly disregard this part. What about the axioms? Set theory comes in many flavors. The main one (to the cries of several people) is $\sf ZFC$, the theory of Zermelo and Fraenkel with the Axiom of Choice.
We can add, or remove, some axioms from the list of axioms which is $\sf ZFC$, but mainly set theorists work within the confines of this theory (and any addition is specified in particular).
So what can we say? We can say that the continuum hypothesis cannot be proved, nor disproved from the axioms of $\sf ZFC$. The proof itself was historically given in two parts, Kurt Gödel showed that we can add another axiom, called today $V=L$, such that $\mathsf{ZFC}+V=L$ proves the continuum hypothesis; and that by adding this axiom we do not introduce an inconsistency (namely, if $\sf ZFC$ didn't prove a false statement, then $\mathsf{ZFC}+V=L$ did not prove one). And two decades later Paul Cohen showed that if $\sf ZFC$ did not prove any false statements then $\sf ZFC+\lnot CH$ did not prove any false statement either.
This shows that $\sf ZFC$ cannot prove, nor disprove the continuum hypothesis. If it could prove it, then Cohen's proof wouldn't work; and if it could disprove it then Gödel's proof wouldn't work. Both proofs do work, to the best of our knowledge, and so it seems that $\sf ZFC$ simply does not prove the continuum hypothesis, unless of course it proves a false statement (in which case we don't want to use these axioms anyway).
Of course, throughout the entire process we assume that $\sf ZFC$ is consistent, otherwise what's the point? And therefore it has a model, namely a particular structure interpreting the relation $\in$ in such way that all the axioms of $\sf ZFC$ are true in that structure. And of course, in a given structure the continuum hypothesis is either true, or it is false. Because in a given structure every sentence is either true, or false (but not both!).
The difficulty, I find, comes from understanding that set theory, like any other theory, has different models. Whether or not there is one intended universe that we care about is irrelevant from this point of view. The theory itself has different models, and within each different statements might be true or false. Statements like the continuum hypothesis.
So what about the two reasons that you gave? Well, neither quite exactly is the reason that the continuum hypothesis is unprovable, but both are true.
First of all, what does it mean that we "find the cardinality of a set"? I can write down a simple definition of a set. Now this definition is interpreted in different models of set theory, in some this set is going to be empty, in others non-empty. What is the set? What is its cardinality? We can't "find out" until we find out which model we are using.
This is the situation with the continuum hypothesis. If we know the model we work in, we have a fighting chance of finding out whether or not it is true or false; but since set theory does not have "an intended model", it doesn't have some guideline as to whether or not this statement is true or false.
Secondly, sets which are ineffable and unthinkable, those are all around us. Can you even imagine how does the set $V_\gamma$ where $\gamma=\beth_{\omega_1^{CK}+\omega}$ looks like? It's quite unthinkable. Pretty much anything that you can imagine already happened so far below this set. And yet, it's just a small fragment of a universe of set theory.
Not to mention that as before, we run into difficulties since the set I wrote above is just a definition of a set, and in different models of set theory it will be interpreted differently. So even if you can imagine it in one model of set theory, you might not be able to imagine it in another. Not in "vivid details" like you can imagine the natural numbers.
And for that matter, can you even imagine a difference between $\Bbb Q$ and $\Bbb{R\setminus Q}$? Both have the same properties as ordered sets, but they are not of the same cardinality. Imagination is overrated when it comes to infinite sets, and even more so when it comes to uncountable sets. And in set theory, countable sets are just the tip of the iceberg.
So why is the continuum hypothesis unprovable? Well, because we chose a weak theory (namely $\sf ZFC$). But that's a good thing. It's good when you theory is weak, because it would require less justifications (philosophically or mathematically) as to why it is true.
I think that somewhere in the early 1960s it was expected that Gödel's axiom, $V=L$ will be accepted into the set theoretical canon. But it didn't, and thank goodness too. Because Cohen's proof opened up a huge world of interest in unprovable statements, that the majority of which are incompatible with $V=L$.