Infinite sets with cardinality less than the natural numbers

No there are none. If $A$ has cardinality of at most the natural numbers, we may assume that it is a subset of the natural numbers.

One can show that a subset of the natural numbers is either bounded and finite, or unbounded and equipotent to the natural numbers themselves.

Every set $X$ that has cardinality at most that of $\mathbb{N}$, i.e. such that $X$ injects into $\mathbb{N}$, is either finite or has the same cardinality as $\mathbb{N}$ by the argument in Asaf's answer.

Moreover assuming the Axiom of Choice or even just a weak fragment thereof, every set $X$ that does not have cardinality at least that of $\mathbb{N}$, i.e. such that $\mathbb{N}$ does not inject into $X$, is finite by the argument in user103567's answer.

However, perhaps it is worth pointing out that it is consistent with $\mathsf{ZF}$ set theory, which is like the usual theory $\mathsf{ZFC}$ but without the Axiom of Choice, that there are infinite sets $X$ whose cardinality is incomparable with that of $\mathbb{N}$; i.e. neither $X$ nor $\mathbb{N}$ injects into the other.

Such pathological sets would be infinite and Dedekind finite: see here for the definitions.

Let $A$ be a infinite set.

Choose a element $a_1\in A$, then $A-a_1$ is still infinite.

....

Choose a element $a_n\in A$, then $A-a_1-a_2-\dots-a_n$ is still infinite.

Since $A$ is infinite, we can do this infinite times(countable), then we get a subset of $A$,i.e.,$\{a_1,\dots,a_n,\dots\}$, which shows that we have constructed a one to one mappi