Theory of a class of finite L-Structures Exercise 2.1.2.3. in "A Course in Model Theory" by Ziegler, Tent states:
Assume that $\mathcal{C}$ is a class of finite $L$-structures containing only finitely many structures of size $n$ for each $n\in \omega$. Then the infinite models of Th$(\mathcal{C})=\{\varphi\;|\;\mathfrak{A}\vDash\varphi\text{ for all }\mathfrak{A}\in\mathcal{C}\}$ are exactly the models of
\begin{align}
\text{Th}_{\text{a}}(\mathcal{C})= \{\varphi\;|\;\mathfrak{A}\vDash\varphi\text{ for all but finitely many }\mathfrak{A}\in\mathcal{C}\}
\end{align}
I did not manage to find a proof of this statement.
(I assume that $\mathcal{C}$ is infinite, since otherwise only finitely many structures in $\mathcal{C}$ would not satisfy $\theta=\neg\exists x_1\cdots \exists x_n\;  \bigwedge_{i<j} \neg x_i\dot{=} x_j$, i.e. $\theta\in$ Th$_{\text{a}}(\mathcal{C})$ while for any infinite $\mathfrak{A}\vDash\text{Th}(\mathcal{C})$ we have $\mathfrak{A}\not\vDash\theta$.)
When I try to start with an infinite model of $\text{Th}(\mathcal{C})$ I see no reason why it should satisfy any $\varphi\in\text{Th}_{\text{a}}(\mathcal{C})\setminus\text{Th}(\mathcal{C})$ since not every infinite model of Th$(\mathcal{C})$ has to be constructed in way so that elements satisfying $\varphi$ are included.
Counter example to make it more clear: $\mathcal{C}:=\{\mathfrak{A}_n:=(\{1,...,n\}, L)\;|\; n=2,3,4...\}$ with $L=\{R\}$ and $R$ being an $4$-ary relation symbol interpreted as $R(a_1,a_2,a_3,a_4)$ if and only if the informal condition $a_1<a_2<a_3<a_4$ and $a_1,...,a_4<6$ is true. Then
\begin{align}
\mathfrak{A}_n\vDash \varphi:=\exists x_1\exists x_2\exists x_3\exists x_4 R(x_1,x_2,x_3,x_4)\;\text{ with }\;
n>3
\end{align}
and, therefore, $\varphi\in\text{Th}_{\text{a}}(\mathcal{C})$. 
Then $\mathfrak{B}=(\{1,3,5,7,...\},L)$ is an infinite model of Th$(\mathcal{C})=\{\exists x_1\exists x_2\;\neg x_1\dot{=} x_2\}$, but $\mathfrak{B}\not\vDash\varphi$.
Where is my misconception?
 A: First of all, note that in your example $Th(\mathcal{C})$ is more complicated than you claim (you write that it is just $\{\exists x_1, x_2(x_1\not=x_2)\}$): for example, the sentence $$\varphi=\mbox{"If there are at least four elements, then there is an instance of $R$"}$$ is true in every structure in $\mathcal{C}$. And this is first-order expressible; more generally, the sentence $\psi_n$="there are at least $n$ elements" is first-order expressible for each natural number $n$. Now as you observe we have $\mathfrak{B}\not\models\varphi$, so in fact $\mathfrak{B}$ is not a model of $Th(\mathcal{C})$.

This observation is also the key to proving the statement in question. For $\varphi\in Th_a(\mathcal{C})$, write "$height(\varphi)$" for the least $k$ such that every  element of $\mathcal{C}$ with at least $k$-many elements satisfies $\varphi$. Note that if $\varphi\in Th_a(\mathcal{C})$ then $height(\varphi)$ exists. The key point is now:

For a sentence $\varphi$, we have $\varphi\in Th_a(\mathcal{C})$ iff $(\psi_{height(\varphi)}\implies\varphi)\in Th(\mathcal{C})$.

(The left-to-right direction is trivial; for the right-to-left direction, what assumption about $\mathcal{C}$ do we need to use?)
From this, do you see how to conclude that $M\models Th(\mathcal{C})\implies M\models Th_a(\mathcal{C})$ for $M$ infinite?
