A set is infinite iff it is equivalent to a proper subset of itself

A certain section of the chapter "The Axiom of Choice" on "Naive Set Theory" got me confused:

"[...] The assertion is that a set is infinite if and only if it is equivalent to a proper subset of itself. The "if" we already know; it says merely that a finite set cannot be equivalent to a proper subset. To prove the "only if," suppose that $X$ is infinite, and let $v$ be a one-to-one correspondence from $\omega$ into $X$. If $x$ is in the range of $v$, say $x=v(n)$, write $h(x)=v(n^{+})$; if $x$ is not in the range of $v$, write $h(x)=x$. It is easy to verify that $h$ is a one-to-one correspondence from $X$ into itself. Since the range of $h$ is a proper subset of $X$ (it does not contain $v(0)$), the proof of the corollary is complete. The assertion of the corollary was used by Dedekind as the very definition of infinity."

(The corollary would follow from "every infinite set has a subset equivalent to $\omega$".)

The question is: how can $h$ be a one-to-one correspondence from $X$ into itself if its range does not contain an element (namely, $v(0)$) that is in $X$?

Note that the word "equivalent" requires context, literally it would just be interpreted as "satisfying some equivalence relation", but what the nature of this relation is not automatically understood.

In the context of cardinality it means to have a bijection. Namely an infinite set has a bijection with a proper subset of itself. Of course this requires the axiom of choice (or rather a small fragment of it) to hold. This characterization is known as Dedekind-infinite. Without the axiom of choice it is consistent that there are sets which are infinite (in the sense that they are not with bijection with any finite ordinal), but they are not Dedekind-infinite.

For example note that $f(n)=n+1$ is a bijection from $\omega$ into $\omega\setminus\{0\}$. It shows that $\omega$ is a Dedekind-infinite set. For the more general case see my answer for Equivalent characterisations of Dedekind-finite proof.