Let's we prove the following theorem.
Theorem
An ordinal $\nu$ is a natural if and only if there is no injection of $\nu$ into $X$ in $\mathscr P(\nu)\setminus\{\nu\}$.
Proof.
Let's we assume there no exists an injection $f$ of $\nu$ over any $X$ in $\mathscr P(\nu)\setminus\{\nu\}$. So if $\nu$ was not a natural the there would be exists $A$ in $\mathscr P(\nu)$ with no maximum element so that let be $\alpha$ the corresponding initial ordinal: this ordinal exists independently by AC since $A$ is a set of ordinals so that it is well ordered! Now if $\alpha$ was in $\nu$ then by transitivity the inclusion $$ \alpha\subseteq\nu $$ would hold and thus by the above hypothesis it must be an equality: we conclude in this way that the inequality $$ \nu\le\alpha\tag{1}\label{1} $$ holds. So by ineq. \eqref{1} we observe that the inequality $$ |\nu|\le|\alpha|=\alpha $$ holds but by the inclusion $$ A\subseteq\nu $$ also the inequality $$ \alpha=|A|\le|\nu| $$ holds so that we argue the equality $$ \alpha=|\nu| $$ holds: thus if $\alpha$ is initial then by ineq. \eqref{1} we conclude the equality $$ \alpha=\nu\tag{2}\label{2} $$ Now by hypothesis $A$ has not maximum element so that if it is equipotent to $\alpha$ then by eq. \eqref{2} we argue that $\nu$ has not maximum element so that it is a limit ordinal and thus surely the inequality $$ \omega\le\nu $$ holds since we supposed $\nu$ is not a natural so that it is not empty. Now the function $$ f:=\big\{(n,m)\in\omega\times\omega:m=n+1\big\} $$ is a bjection over $\omega_+$ so that the function $$ g:=\big\{(\delta,\lambda)\in\nu\times\nu:\lambda=f(\delta)\text{ if $\delta\in\omega$},\lambda=\delta\text{ otherwise}\big\} $$ is an bjection of $\nu$ onto $\nu\setminus\{\emptyset\}$ and this is impossible by original hypothesis: so we conclude that $\nu$ is an natural.
Finally if $\nu$ is a natural number then by this lemma we conclude that there is no injection of $\nu$ over $X$ in $\mathscr P(\nu)\setminus\{\nu\}$.
So first of all I ask if the above theorem holds (I elaborate it by myself!) and so if I well proved it; moreover, I would like to know if it is possibile the implication I did not prove directly whithout induction -or recursion: could someone help me, please?
For natural numbers, you can get an essentially identical proof to the one by induction by saying 'let $n$ be the least natural number such that $n$ has an injection onto a proper subset of itself. Then use the same argument as in the linked proof to show that $n-1$ has one'.
You can get a different proof if using a different equivalent definition of finiteness. For example, the complementary version of Tarski-finiteness. A set $X$ is finite if and only if every non-empty $Y\subseteq\mathcal{P}(X)$ has a $\subseteq$-minimal element.
Assume $X$ is Tarski-finite and $f\colon X\to X$ is injective onto a proper subset of $X$. Let $$ Y=\{A\subseteq X\colon f(A)\subsetneq A\} $$ Then $Y$ is nonempty because $X\in Y$. Let $A_0$ be a $\subseteq$-minimal element of $Y$. Then $f(A_0)\subsetneq A_0$. Let $t\in A_0\setminus f(A_0)$. Let $B=A_0\setminus \{t\}$.
Since $t\not\in f(A_0)$ then $f(A_0)\subseteq B$. Therefore $f(B)\subseteq B$. Let $u=f(t)$. Then $u\in B$, and since $f$ is injective, there is no $s\in B$ such that $f(s)=u$. This shows that $f(B)\subsetneq B$. But then $B\in Y$, contradicting the $\subseteq$-minimality of $A_0$.
Your proof for infinite ordinals is fine. You can summarize to say there is a class function $F$ defined on all ordinals by $F(n)=n+1$ for $n<\omega$ and $F(\alpha)=\alpha$ for $\alpha\ge\omega$. Then for all $\alpha\ge\omega$, $F\rvert_\alpha$ is an injective function from $\alpha$ onto a proper subset of $\alpha$.
Edit: If $X$ is a finite set with $|X|=n$, the argument above shows that $$ X\supsetneq f(X)\supsetneq f^2(X)\supsetneq f^3(X)\supsetneq\ldots $$ Since $X$ has $n$ elements and each step deletes at least one element, after at most $n$ iterations we get the empty set and that leads to a contradiction since $\emptyset\supsetneq f(\emptyset)$ is impossible.
This is also a proof by induction, but hopefully makes it more transparent what's going on. Since each application of $f$ contracts $X$, starting with a finite set, eventually the process drains $X$. If $X$ is (Dedekind) infinite it's possible to repeat such a process indefinitely many (finite) number of steps and still remain with a set which is 'as large' as the original.