Given a theory $S$, $M(S)$ denotes the class of models of $S$, that is, if $A\in M(S)$, then $A\vDash S$.
The theorem goes like this: Given two theories $S$ and $T$ of the same first order language such that $M(S)\cap M(T)=\emptyset$, there exists $\theta\in S$ and $\phi\in T$ such that no models of $\theta$ is a model of $\phi$.
To prove the contrapositive statement, suppose that for every pair of $\theta\in S$ and $\phi\in T$, there exists a model $A$ such that $A\vDash\theta\land\phi$.
Let $\Pi\subset S\cup T$ be a finite subset. If I can show that $\Pi$ has a model, then by compactness theorem we can conclude that $S\cup T$ has a model, which finishes the proof.
If $\Pi\subset S$ or $\Pi\subset T$, then the result follows immediately (since $M(S)\neq\emptyset$ and $M(T)\neq\emptyset$). So suppose $\Pi=\{\theta_1,...\theta_m,\phi_1,...\phi_n\}$ with each $\theta_i\in S$ and each $\phi_i\in T$. Then for each pair of $\theta_i$ and $\phi_j$ there is a model $A_{i,j}$ such that $A_{i,j}\vDash\theta_i\land\phi_j$.
Here is where I'm stuck, I want to construct a model $A$ that satisfies $\theta_i\land\phi_j$ for each $i\leq m$ and $j\leq n$. But I'm not sure how/if am I able to do that.
To prove the contrapositive statement, suppose that for every pair of $\theta\in S$ and $\phi\in T$, there exists a model $A$ such that $A\vDash\theta\land\phi$.
Suppose for a contradiction $S\cup T$ is not satisfiable. Then by compactness theorem there exists a finite subset $S_0\cup T_0\subset S\cup T$ that is not satisfiable. Let $S_0\cup T_0$ be such, with $S_0=\{\theta_1,...,\theta_m\}\subset S$ and $T_0=\{\phi_1,...,\phi_n\}\subset T$. Then $S_0\cup T_0=\{\theta_1,...,\theta_m,\phi_1,...,\phi_n\}$.
Let $\theta=\theta_1\land...\theta_m$, $\phi=\phi_1\land...\phi_n$. Then by definition of theory, $\theta\in S$ and $\phi\in T$. By our hypothesis, there exists a model $A$ such that $A\vDash\theta\land\phi$. Hence $A\vDash\theta_1\land...\land\theta_m\land\phi_1\land...\land\phi_n$. Hence $A\vDash S_0\cup T_0$, which contradicts that $S_0\cup T_0$ being not satisfiable.
Hence $S\cup T$ is satisfiable. So $M(S)\cap M(T)\neq\emptyset$.