Using the Axiom of Choice we know that any set $X$ is equipotent to an unique initial ordinal $|X|$ that we call the cardinal of $X$, so we can organize the elements of $X$ in a transfinite succession, that is $X=\{x_0,...,x_n,...\}$, and so for any function $\phi:X\rightarrow Y$ we can define the function
$$ \Phi:|X|\owns\alpha\rightarrow\phi(x_\alpha)\in Y $$
therefore for any function $f$ we can define the ordinal $\mu_f=\min\{0\le\beta<|X|:\Phi(\beta)\notin f\}$ and so the operation
$$ \mathbf{G}(f):=\Phi(\mu_f) $$
that through the transfinite recursion theorem allow to claim that there exist an operation $\mathbf{F}$ such that for any ordinal $\alpha$ it result that
$$ \mathbf{F}(\alpha)\equiv\mathbf{G}(\mathbf{F}|_\alpha)=\Phi(\mu_{\mathbf{F}|_\alpha})=\phi(x_{\mu_{\mathbf{F}|_\alpha}}). $$
Well now through $\mathbf{F}$ and through the Axiom Schema of Comprehension we selected a set $Z=\{x_{\mu_{\mathbf{F}|_0}},...x_{\mu_{\mathbf{F}|_n}},...\}\subseteq X$ such that $\phi|_Z$ is injective and such that $\phi(Z)=\mathscr{ran}\phi:=\phi(X)$ and so it result that $|\phi(X)|\le|X|$.
Is my prove correct? if not, how demonstrate the assertion? maybe the assertion is false?
Could someone help me, please?
I do not understand the use of transfinite recursion in this case.
Once you have fixed a well-ordering on $X$ you are more or less done. For every $y\in\phi(X)$ simply define $\psi(y)$ to $x_\alpha$ with the minimum index such that $\phi(x_\alpha)=y$.
No recursion is needed. In fact, depending on how you formulate the axiom of choice, you might not even need to well-order $X$ at all. Simply note that for every $y\in\phi(X)$ the set $X_y=\{x\in X\mid\phi(x)=y\}$ is not empty, so there is a choice function, and since $X_y\cap X_{y'}=\varnothing$ if and only if $y\neq y'$, such a choice function is in fact injective.
In fact, in some situations one states the axiom of choice as "every surjection has an injective inverse", in which case you have to do exactly nothing to prove the claim.
In none of these cases we need to use transfinite recursion, though. Except if we want to somehow first prove that $X$ can be well-ordered and then use the well-ordering to define a choice function.