Let $\mathfrak{A}$ be an $L$-structure with domain $A$. If $\Sigma$ is a finite set of formulas in $L(A)$, how can I prove that $\Sigma$ is realized in $\mathfrak{A}$ iff it is consistent with $Th(\mathfrak{A})$?
Here is my attempt:
If $\Sigma \subset L(A)$-Form is such that $\mathfrak{A} \vDash_{L(A)} \Sigma[\overline{a}]$ (that is, $\mathfrak{A} \vDash_{L(A)} \varphi[\overline{a}]$ for every $\varphi \in \Sigma$) for some $\overline{a} \in A^n$, then there is an $\overline{a'} \in A^k$ $(k\geq n) $ because $\Sigma$ is finite and therefore for every $c_x \in \varphi$ for some $\varphi \in \Sigma$ and $c_x$ a constant symbol in $L(A)$ for an element $x \in A$ we can complete $\overline{a}$ to $\overline{a'}$ such that $\mathfrak{A} \vDash_L \varphi[\overline{a'}] $ and since $\mathfrak{A} \vDash_L Th(\mathfrak{A})$ $\Sigma \cup Th(\mathfrak{A})$ has a model ($\mathfrak{A}$) and is therefore consistent. The $\Leftarrow$ follows the same reasoning.
My question is: Why can't we say that $\mathfrak{A} \vDash_{L(A)} Th(\mathfrak{A})$ and avoid the finiteness condition?
If I've correctly deciphered your one-sentence proof of the "only if" direction, it's correct, and it works also for infinite sets $\Sigma$. The "if" direction, however, needs finiteness. Your assertion that it "follows the same reasoning" seems to be wrong.
Here's an example showing that the "if" direction can fail when $\Sigma$ is infinite. Let $\mathfrak A$ be the structure consisting of the natural numbers, with only the equality relation. Let $\Sigma$ be the set of all $L(A)$-formulas of the form $\neg(x=n)$ for $n\in A$. Then $\Sigma$ is consistent with Th$(\mathfrak A)$, because a deduction of a contradiction would involve only finitely many formulas from $\Sigma$ whereas each finite subset of $\Sigma$ is realized in $\mathfrak A$ and therefore consistent with Th$(\mathfrak A)$. But $\Sigma$ is not realized in $\mathfrak A$.
EDIT: To prove the "if" direction in the case of finite $\Sigma$, first form the conjunction $\sigma$ of all the formulas in $\sigma$. (This is where I used that $\Sigma$ is finite, so that $\sigma$ is a first-order formula.) If $x_1,\dots,x_n$ are the free variables in $\sigma$, consider the sentence $\exists x_1\dots\exists x_n\,\sigma$, which I'll abbreviate as $\xi$. If this sentence $\xi$ were false in $\mathfrak A$, then its negation would be in Th$(\mathfrak A)$, and it would follow easily that $\Sigma$ is inconsistent with Th$(\mathfrak A)$, contrary to hypothesis. So $\xi$ is true in $\mathfrak A$, and this means exactly that $\Sigma$ is realized in $\mathfrak A$.