Q:

Adding enough sets to get a model of ZF

Working in (first or second-order) ZF, suppose we started with an arbitrary set $A$ containing $\omega$ and wanted to add enough things to $A$ to get a model, not necessarily minimal, of (first or second-order) ZF.  Is there anything wrong with the following brute force approach? 
$T_0 = A$
$T_{\alpha+1} = trcl(T_{\alpha}) \cup\hspace{2mm} \{f[X]\hspace{1.5mm} |\hspace{2mm} X \in T_{\alpha} ,f:X \rightarrow V  \} \cup \{ \cup(x) \hspace{1.5mm} |\hspace{2mm} x \in T_{\alpha}\} \cup  \{ \wp(x) \hspace{1.5mm} |\hspace{2mm} x \in T_{\alpha}\}$
$T_{\alpha} = \bigcup_{\beta<\alpha} T_\beta$ for $\alpha$ limit. 
Let $T = \bigcup_{\alpha \in \textrm{ORD}} T_{\alpha}$ 

Then $T \models$ ZF. 
Proof: $T$ is transitive, and that suffices for $T$ to model Extensionality and Foundation.  
$T$ models Replacement because whenever $X \in T$, there is some $\alpha$ such that $X \in T_{\alpha}$. So given $f: X \rightarrow V$, then $rng(f) \in T_{\alpha+1} \subset T$. 
$T$ models Powerset because whenever $x \in T$, there is some $\alpha$ such that $x \in T_{\alpha}$, and then $\wp(x) \in T_{\alpha+1} \subset T$.
$T$ models Union because whenever $x \in T$, there is some $\alpha$ such that $x \in T_{\alpha}$, and then $\cup(x) \in T_{\alpha+1} \subset T$.
$T$ models Infinity by assumption that $\omega \in A$.  
Separation and Pairing should follow from Replacement (and Powerset in the case of Pairing). 

A:

Well, this "works" in that it gives you a (class) model of ZF, but it just gives you the entire universe $V$.  Indeed, as soon as $T_\alpha$ contains a nonempty set, $T_{\alpha+1}$ will contain every singleton set, since every singleton is the image of some function on any nonempty set.   Then taking unions will give that $T_{\alpha+2}$ is all of $V$.
However, it is not necessary to consider arbitrary functions as you have done.   Instead of functions $f:X\to V$, you can consider just functions $f:X\to T_\alpha$.  Then your construction still works (since any function from a set to $T$ will have image contained in some $T_\alpha$), and if you start with a set $A$ of rank $\alpha$, it is not too hard to show that your model $T$ will be $V_\kappa$ where $\kappa$ is the least inaccessible cardinal greater than $\alpha$, or all of $V$ if no such $\kappa$ exists.   Indeed, your construction is basically constructing the smallest Grothendieck universe containing $A$.