Let $L$ be a language and let $T$ be an $L$-Theory. $M\models{T}$ is said to be locally finite if for any given finite subset $X$ of $M$, there is a finite substructure $A$ of $M$ s.t. $X\subseteq{M}$. The theory $T$ is locally finite if all of its models are locally finite.
I wish to show that $T$ is locally finite iff $T_{\forall}$ is locally finite.
Showing the RHS implies LHS is straightforward. I'm however having trouble showing that LHS implies RHS. My attempts so far have been along the following lines:
Let $A\models{T_{\forall}}$. Now let $X\subseteq{A}$ be finite. We know that there is an $M$ such that $M\models{T}$ and $A$ is a substructure of $M$. Thus there is a finite substructure $B$ of $M$ that contains $X$. But the problem is that $B$ may not lie inside of $A$. I would like to use a sort of "general nonsense" argument to create a copy of $B$ inside of $A$ (assuming of course that $|B|<|A|$). However I'm having trouble setting up the required isomorphism. Is this the correct way to approach this problem? Or am I missing some details?
Instead of taking just any finite substructure of $M$ containing $X$, take the substructure generated by $X$ (the smallest substructure containing $X$). This is contained in $A$ because $A$ is also a substructure.