My professor taught us this proof that the set of integers is infinite:
Suppose $Z$ is not infinite, therefore there is a number $m\in N$ such that exists a bijection $f:Im\rightarrow Z$. That means $f^{-1}:Z\rightarrow Im$ is also a bijection, but if we pick $n \in N$ such that $n>m$, $In\subset Z$, the restricted function $f^{-1}|_{In} : In\rightarrow Im$ is still an injection which is a contradiction, since there can't be an injection with a domain that has more elements than the codomain, therefore Z is infinite.
However I can't understand why we can assume $In\subset Z$ when we already said $Z$ has $m$ elements and $In$ has more than $m$ elements.
Ps: Sorry about my english, it's not my first language.
The deduction
$$f^{-1} : \mathbb Z \to I_m \ \text{ exists and is a bijection } \Longrightarrow f^{-1}|_{I_n} : I_n \to I_m \ \text{ exists and is an injection }$$
itself is sound.
The nub of your complaint is what leads to the contradiction: there cannot be such an $f^{-1}$ nor such an $f$.
In other words, you are demonstrating the logical inconsistency in a different way that the argument is written. It is not surprising that you can deduce a logical contradiction from one step of the chain of reasoning. The chain of reasoning is written to ultimately demonstrate such a contradiction. Either way (the way the argument is written or the logical inconsistency you found), we arrive at the desired contradiction and necessarily conclude that the starting premise is false.