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.
Let $\frak p$ and $\frak q$ be cardinals such that $\frak p<q$.
- $\frak q$ is the $1$-successor of $\frak p$ if whenever $\frak m$ is such that $\frak p\leq m\leq q$ then $\frak p=m$ or $\frak q=m$.
- $\frak q$ is the $2$-successor of $\frak p$ if whenever $\frak m$ is such that $\frak p < m$ then $\frak q\leq m$.
- $\frak q$ is the $3$-successor of $\frak p$ if whenever $\frak m$ is such that $\frak m < q$ then $\frak m\leq p$.
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 $\aleph_0$.