I think for the one-one mapping, to $(A\backslash x_1)\cup Y$, from $X$, there is one element which must be borrowed from $Y$, but the author writes all of $Y$ is mapped to itself.
In mapping one-one from the finite subset of $X$, $A=\{x_1, x_2, \ldots, x_n\}$, the image of the function needs the $n$th element to be mapped to a member of $Y$, doesn't it?
Shouldn't one just say the map borrows some element from $Y$, shifting all of $Y$ by one as well?
The second to last paragraph of section 4, of M. Rosenlicht's Intro to Analysis, reads as follows:
It is easy to show that a set $X$ is infinite if and only if it may be put into one-one correspondence with a proper subset of itself.
To do this, note first that if $X$ is finite then any proper subset has a smaller number of elements, whereas two finite sets in one-one correspondence must have the same number of elements.
This proves the "if" part.
On the other hand, if $X$ is infinite then there exist distinct elements $x_1, x_2, x_3, …$ in $X$.
The complement of $x_1, x_2, x_3, \ldots$ in $X$ is a subset $Y$, so that
$$
X = \{x_1, x_2, x_3, \ldots\} \cup Y \textrm{and} \{x_1, x_2, x_3, \ldots\} \cap Y = \emptyset.
$$
A one-one correspondence between $X$ and its proper subset $\{ x_2, x_3, x_4, \ldots\} \cup Y$ is given by the function which sends each $x_n$ into $x_{n+1}$ and each element of Y into itself.
This proves the "only if" part, completing the proof.
I believe the distinct elements goes along with the Axiom of Choice, as used in answers to the other questions I found on this topic.
Best Answer
You are using the Axiom of Choice to select (and denumerate) a countably infinite subset of $X$. You can always do that because $X$ is infinite. Let's say $S= \{x_1, x_2, \ldots \}.$ Anything that's left over is defined as being in $Y$; in other words, $Y$ is defined by $X \setminus S$. You're then mapping the countably infinite set $S ~1-1$ into a proper subset of itself and leaving $Y$ alone, to achieve a $1-1$ map from $X \to X \setminus \{x_1\}$.