If $\operatorname{Mod}(T_1 \cup T_2) = \emptyset$ then for some $\sigma$, $T_1 \vDash \sigma$ and $T_2 \vDash \neg \sigma$ Problem description: if $T_1$ and $T_2$ are theories such that $\operatorname{Mod}(T_1 \cup T_2) = \emptyset$, then there is a $\sigma$ such that $T_1 \vDash \sigma$ and $T_2 \vDash \neg \sigma$. 
I don’t understand how to approach this. Is it even true if I’m not given that $T_1$ and $T_2$ are consistent? 
Hints only please; no full solutions. Thanks! 
 A: If $T_1 \cup T_2$ is inconsistent, then $T_1 \cup T_2 \vdash \bot$, where $\bot$ is some falsehood such as $\exists x\,(x\neq x)$. Let $S_i$ be all the sentences of $T_i$ that occur in the proof of $\bot$, $i = 1,2$. Then $S_1 \cup S_2\vdash \bot$. 
Now let $\sigma_i = \bigwedge_{\varphi\in S_i} \varphi, i = 1,2$. Then 
$$
\vdash (\sigma_1 \land \sigma_2) \to \bot,
$$
i.e.
$$
\vdash \sigma_1 \to \neg\, \sigma_2.
$$
Note that $T_i \vdash \sigma_i, i = 1,2$.
A: Theorem (Compactness). A theory $T$ is consistent iff every finite $S\subset T$ is consistent.Theorem (Henkin). A theory is consistent iff it has a model. So if $T_1\cup T_2$ has no model, then it is inconsistent.
Suppose $T_1$ and $T_2$ are consistent. By Compactness, we have $\neg$ Con $(T_1\cup U)$ for some finite $U\subset T_2,$ and since Con $(T_1)$ we have $U\neq \phi.$  Take $u \in U$ and let $W=U\backslash \{u\}.$ Let $\sigma$ be $[(\land_{w\in W})\implies \neg u].$
I will leave the cases $\neg$ Con$(T_1)$ and $\neg$ Con$(T_2)$ to someone else.
