Let X be a finite set and Y a subset of it.
If $X=\emptyset$ then Y is finite.
If $X \neq \emptyset$, then we have, by definition of a finite set, the following bijection: $$ f:I_n\rightarrow X $$ Where $I_n = \{1,...,n\}$.
Since $Y\subset X$, $f^{-1}(Y) \subset I_n$ and $f|_{f^{-1}(Y)}:f^{-1}(Y)\rightarrow Y$.
So If I'm able to show that $f^{-1}(Y)$ is finite, we'll have: $$ g:I_k\rightarrow f^{-1}(Y) $$ bijective, and I can compose functions to get: $$ h:I_k\rightarrow Y $$ concluding that Y is finite.
Let's suppose that $X = I_n$ and $Y\subset X$.
For $n = 1$ we have $I_n = \{1\}$, wich implies that $Y=\{1\}$ or $Y=\emptyset$, in both cases $Y$ is finite.
Now let's suppose the theorem is valid for an arbitrary n, and therefore, let's try to prove that it's valid for $n+1$.
Since we have $Y\subset I_{n+1}$, we can have $Y=I_{n+1} \rightarrow$ Y is finite, and $Y \neq I_{n+1} \rightarrow \exists a \in I_{n+1} - Y$.
Let's now define a bijection k:
$$ k:I_{n+1}\rightarrow I_{n+1} $$
where $k(a) = n+1$.
Therefore:
$$ k|_Y:Y\rightarrow k(Y)\subset I_{n+1} - \{n+1\} = I_n $$
And by our induction hypothesis, we can see that $k(Y)$ is finite.
If $k(y)$ is finite and $k|_y$ is a bijection, we can compose functions, as I said in the beginning, to see that Y is finite
My question is... I needed to suppose that $X=I_n$ to prove it. It's ok to suppose that? And if so, why?