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In ZF, an injection $h:\omega \to X$ yields an injective nonsurjective self-map by shifting $h(n)$ to $h(n+1)$ and fixing the complement. Conversely, if an injection $g:X\to X$ omits $a$, the natural-number recursion theorem defines $h(n)=g^n(a)$. Any repetition would, after cancellation, put $a$ in the image of $g$. Thus $h$ is injective. The argument uses no axiom of choice, even when some infinite sets are Dedekind-finite sets.

If $X$ is a Dedekind-finite set, its set of finite repetition-free sequences is also Dedekind-finite. An injective sequence of distinct finite lists would, by countable union of explicitly ordered finite lists without choice, either produce an injection $\mathbb{N}\to X$ or use only finitely many entries. The latter possibility is impossible because a fixed finite pool supports only finitely many repetition-free lists. If $X$ is infinite, the one-entry lists also show that the resulting set is infinite.