I so stuck with a problem of set theory. But first a recursive definition:
Define $R_0=\emptyset$
If $R_\alpha$ is defined, then $R_{\alpha+1}=\mathcal{P}(R_\alpha)$ (the power set).
For a limit ordinal $\gamma$, if $R_\alpha$ is defined for all $\alpha<\gamma$, then define $R_\gamma=\displaystyle\bigcup_{\alpha<\gamma} R_\alpha$
We define $\text{BF}=\displaystyle\bigcup_{\alpha\in\text{OR}} R_\alpha$ (the class of well founded sets).
Next, my problem
Take $A\subseteq \text{BF}$ a proper transitive class such that $(A,\in)\models\text{ZF}$. Prove that $A$ is almost universal.
I think that the exercise is false because if it was true, then we could conclude that the strongly inaccessible cardinals doesn't exists. This because if $\kappa$ is strongly inaccessible then $R_\kappa$ satisfies the hypothesis but $R_\kappa$ isn't almost universal. I really appreciate any hint or/and suggestion.
Edit: my counterexample is wrong. But, then, how can I solve the exercise?
Since $A$ is a transitive proper class model of ZF, then for any ordinal $\alpha,$ we have $R_\alpha\cap A \in A.$ This follows from the absoluteness of the rank function. $R_\alpha \cap A$ is just $A$'s version of $R_\alpha,$ the sets of rank less than $\alpha.$
Now to see that $A$ is almost universal, consider any set $B\subseteq A.$ Then, since $B$ is a set, for some sufficiently large $\alpha$ we have $B\subseteq R_\alpha,$ and hence $B\subseteq R_\alpha\cap A\in A.$