While thinking about the Continuum Hypothesis, I stumbled across a way of thinking about it that seems to intuitively make sense to me. But, being that I'm not a mathematician and Gödel/Cohen together have shown that the $\sf{CH}$ is independent of the $\sf{ZFC}$ axioms, I understand that this is $99.999999999999\%$ likely to be wrong. Anyway, here is my thinking, and if possible, I'd like to know where it is that I went awry:
Idea:
The assumption underlying my "proof" is that the number of natural numbers you need to "index" a set corresponds to its cardinality. Therefore, if you have a set $A$ with a cardinality $a$, and a set $B$ with a cardinality $b$, and you need $i$ indices to index $A$, and $j$ indices to index $B$, and you can show that there exists no number of indices between $i$ and $j$ that would generate a different cardinality than either $a$ or $b$, then there must be no cardinalities between $a$ or $b$.
So, let me show you what I mean by indices. The members of the set of natural numbers need only $1$ natural number to index them. Duh. $1$ gets labeled by $1$, $2$ by $2$, etc. So, a set with cardinality $\aleph_0$ only needs $1$ index. We also know that ordered $n$-tuples (with finite $n$) also are countable, requiring only $1$ index to label them. What is the next possible number of indices you could require? $\aleph_0$. The very next possible number of indices that would generate a different cardinality is $\aleph_0$. If you have $\aleph_0$ indices, each index being a natural number, the set of all such objects would basically be an ordered $\aleph_0$-Tuple. The cardinality of such a set is that of the continuum. Therefore, since there is no possible number of indices to label members of sets between finite natural numbers and $\aleph_0$, there are no cardinalities between $\aleph_0$ and the cardinality of the continuum.
Why wouldn't something like this work? It seems like either something outside my assumption is wrong, or my assumption is independent of $\sf{ZFC}$, and it's just one of many possible axioms that $\sf{ZFC}$ could be extended with…
Best Answer
Your underlying assumption that "everything can be indexed nicely" basically translates to the following:
This, however, is provably false in ZFC! It turns out that we can show in ZFC using Koenig's Theorem that $(\aleph_0)^Y$ can never have size exactly $\aleph_\omega$.
In general, when reasoning about infinities, you need to be very careful and precise. Informal arguments like what you've written above - which hinge on appeals to intuition (how do you justify that every set can be "indexed" nicely?) - are more likely to confuse you than to help, until you have learned enough set theory to know when to trust them.
Incidentally, it's worth pointing out that we know that ZFC (assuming it's consistent!) can't prove the Continuum Hypothesis: if you give me a model of ZFC, I can produce (via forcing) a model of ZFC in which the Continuum Hypothesis is false (and another in which it is true).