Suppose you have a set $S$. Let $T$ be the family of all subsets of $S$. We will prove that there is no one-to-one correspondence between $S$ and $T$.
Suppose we have some function from $S$ to $T$, say $f$. Then for each $x\in S$ we have $f(x)$ is some element of $T$, which is a subset of $S$. The element $x$ might be an element of this subset, or it might not.
Now for each $x$, we will paint $x$ blue if $x\in f(x)$, or green if $x\notin f(x)$. Every element of $S$ is now painted blue or green.
Let $G$ be the set of all green elements of $S$. $G$ is a subset of $S$ and since $T$ is the family of all subsets of $S$, $G$ is an element of $T$.
I claim there is no element $g$ such that $f(g) = G$. Let's suppose that there were and see what happens.
Every element of $S$ is painted blue or green. If there were a $g$ with $f(g) = G$, then $g$ would be either blue or green.
If $g$ were green, it would be an element of $G$, the set of all green elements. But we would have painted $g$ green only if we found that $g\notin f(g) = G$. So $g$ cannot be green.
Perhaps $g$ is blue. We would have painted $g$ blue if we found that $g\in f(g) = G$. But $G$ is the set of all green elements, so if $g$ is blue, it is not in $G$, and we would not have painted it blue in the first place. So again we have a contradiction.
Something is seriously wrong. Certainly we can paint the elements as we said. The only plausible error is our claim that there is a $g$ for which $f(g)= G$. So the function $f$ is not in fact a one-to-one correspondence because there is no $g$ for which $f(g) = G$.
We can apply this argument for any proposed function $f$ and show that $f$ is not a one-to-one correspondence. And the argument works for any set $S$.
If you're still puzzled, try doing this construction for a finite set, such as $\{\mathrm{fish}, \mathrm{dog}, \mathrm{carrot}\}$.
Addendum: The argument above proves that there's no one-to-one correspondence between $S$ and $T$, but it's at least conceivable that $S$ is bigger than $T$. But the obvious mapping that takes $x$ to the set $\{x\}$ shows that there is a proper subset of $T$ that is the same size as $S$, so $T$ is at least as big.
Interesting note: We don't need to assume the existence of a bijective function and then show there is a contradiction. Instead we let $f$ be an arbitrary function and then show constructively that it is not a surjection by producing a specific example of an element not in its range. This is a stronger result than the proof by contradiction would have been.
Best Answer
It is consistent without the axiom of choice that there are uncountable sets which cannot be well-ordered, and every infinite set whose cardinality is smaller is necessarily countable. It could also be the case that there are smaller cardinalities, but none which are strictly between $\aleph_0$ and the uncountable set.
Indeed there are three different definitions of successor cardinals when the axiom of choice fails (although the existence of one [for every set] implies choice; there is another which is provable without choice; and a third which is independent).
It is consistent, if so, that there are several successors to $\aleph_0$. It is always true that $\aleph_1$ is the successor of $\aleph_0$, and that it is the minimal aleph above it.
In particular it is consistent that the real numbers form such set.
See, for example, Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
Edit: I have some free time so here are different variants of "successor". This is taken from Jech The Axiom of Choice (p. 163), the original definition is due to Tarski.
Now we see that it is always true that $\aleph_1$ is a $1$- and $3$-successor of $\aleph_0$. However it is consistent that so are the real numbers themselves.
The assertion that $\aleph_1$ is a $2$-successor of $\aleph_0$ is equivalent to every uncountable set $X$ has an injection from $\omega_1$ into $X$. In fact just requiring that $\aleph_0$ has a $2$-successor is enough.
It is consistent that there are several $1$-successors to $\aleph_0$, and that there is a $1$-successor which is not $3$-successor.
For the real numbers it holds that if their cardinal is a $1$-successor of $\aleph_0$ then it is a $3$-successor of $\aleph_0$.
Interestingly, it is consistent that there is a proper class of $1$-successors to $\aleph_0$.