Dedekind infinitude is equivalent to a countable subset

For an injective nonsurjective $f:X\to X$, fix $a\mathrm{\notin }f[X]$ and put $a_n=f^n(a)$. If $a_m=a_n$ with $m<n$, cancel $f$ repeatedly to get $a=f^{n-m}(a)\in f[X]$, impossible. Thus $n\mapsto a_n$ injects omega into $X$.

Given an injection $n\mapsto a_n$, send $\ast$ to $a_0$, each $a_n$ to $a_{n+1}$, and every other point of $X$ to itself. This is a bijection $X\bigsqcup \{\ast \}\to X$. Conversely, restricting any such bijection to $X$ is injective and misses the image of $\ast$, proving Dedekind infinitude.

By the least-nonembedding definition, omega embeds into $X$ exactly when $\omega <h(X)$. Negating gives the asserted Dedekind-finite characterization. For finite $X$, including the empty set, no injective nonsurjective self-map exists, as finite counting also shows.