This is from Fundamentals of Mathematics, Volume I: Foundations of Mathematics: The Real Number System and Algebra, by Behnke, Heinrich, et al.
A mapping $f$ is defined to be one-to-one if it has the property $f\left(x\right)=f\left(y\right)\implies x=y.$ A segment $A_n$ is the set of natural numbers less than or equal to $n.$ The expression $n^\prime$ denotes the successor of $n.$
A one-to-one mapping $f$ of $A_{n}$ into itself [i.e., $f\left(A_{n}\right)\subseteq A_{n}$] is a mapping onto $A_{n}.$
We prove this theorem by the argument from $n$ to $n^{\prime}$. For $n=1$ the assertion is clear, since $1$ is the only element of $A_{1}.$ Now let $f$ be a one-to-one mapping of $A_{n^{\prime}}$ into itself. If $n^{\prime}\ne f\left(x\right)$ for all $x\le n,$ then $f$ induces to a one-to-one mapping of $A_{n}$ into itself, so that by the induction hypothesis $f\left(A_{n}\right)=A_{n}.$ But then we can only have $f\left(n^{\prime}\right)=n^{\prime}$, and consequently $f\left(A_{n^{\prime}}\right)=A_{n^{\prime}}.$ But if $n^{\prime}=f\left(k\right)$ for a number $k\le n$, then by setting
$$ g\left(x\right)=\begin{cases} f\left(x\right) & \text{for }k\ne x\le n\\ f\left(n^{\prime}\right) & \text{for }k=x \end{cases}, $$
we define a mapping $g$ of $A_{n}$ into itself, since $f\left(n^{\prime}\right)\ne n^{\prime}=f\left(k\right)$ follows from $n^{\prime}\ne k$. The one-to-one character of $g$ follows easily from that of $f,$ so that by the induction hypothesis we have $g\left(A_{n}\right)=A_{n},$ and consequently $f\left(A_{n}\right)=A_{n}.$
The induction hypothesis of the proof assumes the existence of a one-to-one mapping of a segment into itself. Is it still necessary to prove that such a one-to-one self-mapping exists?
The book later proves that an onto self mapping of a segment is one-to-one. Thus saying that a self-mapping of a segment is one-to-one is equivalent to saying that it is onto. But, again we assume the existence of an onto self-mapping.
Are we actually assuming the existence of one-to-one self-mappings of segments a priori and continuing on our merry way without ever encountering a contradiction?
No. This argument just shows that such a one to one mapping (should any exist) must also be onto.
(Of course there are many such mappings: $n!$ of them for $A_n$. But you don't need to know that for this argument to be correct.)