I need to prove that, for $M$ a given $L$-structure and $T$ be a theory in the language $L$. Show that if for any finitely generated substructure $M'$ of $M$ there is an embedding of $M'$ into a model of $T$, then there is an embedding of $M$ into the model of $T$.
So, I have that for every $M' \subseteq M$ there is an embedding $h : M' \rightarrow Mod(T)$, so I know that I need to apply compactness, I can do that because I have a finitely generated substructures. I also can see that I need to apply the notion of diagram: The map $h$ is an embedding (from M to N) iff (N,h) is a model of Diag(M ). A diagram is defined as follows. $Diag(M) = {\varphi \in L(M) | \ \varphi \ is \ quantifier \ free \ M_M \models \varphi}$.
Now, I feel like I have all the pieces but i have issues putting them together in their place, I think I need to do something like this:
Let $M' \subseteq M$ a finite model of $M$ and let $h$ be an embedding $h : M' \rightarrow Mod(T)$, then $(Mod(T),h)$ is a model of the $Diag(M)$. Here I apply compactness but I think I'm either skipping a step or missing a decent argumentation of why I can do it. Once I have applied compactness, I think I can use the same argument in reverse to say that there is an embedding from $M$ to $Mod(T)$: so "by compactness" $Mod(T,h')$ is is a model of $Diag(M)$, therefore, $h$ is an embedding from $M$ to $Mod(T)$. I would really appreciate your input on how to improve this proof and give in more details.
If every finitely generated structure of $M$ is embeddable in a model of $T$ then Diag$(M)\cup T$ is finitely consistent. By compactness there is $N\models$ Diag$(M)\cup T$. The model $N$ has signature $L(M)$ and interprets the elements on $M$. This interpretation gives the required embedding of $M$ into $N$. (The model you asked for is $N$ reduced to the signature $L$.)