Why is the reduct theory a theory? My textbook says that if $T\subseteq \mathcal{L}$ is a theory, $\mathcal{L}_0\subseteq \mathcal{L}$, and $T_0:=T\cap\mathcal{L}_0$, then $T_0$ is a theory and called the reduct theory in the language $\mathcal{L}_0$. A theory is defined as a set of sentences (formulas without free variables) that is also deductively closed, so that $T\vDash\alpha \Leftrightarrow \alpha\in T$.
Now say we have a set of sentences $S$ that is not yet a theory because it derives one more sentence: $S\vDash\phi$ and $\phi\notin S$. For all other sentences, the closure property holds. Clearly, $T:=S\cup \{\phi\}$ is a theory. Now pick $\mathcal{L}_0=S$. Then $T_0=T\cap S=S$, but $S$ is not a theory.
Where am I going wrong here?
Edit: I think my error may be here:

Clearly, $T:=S\cup \{\phi\}$ is a theory.

After all, $T$ could now be able to derive yet more sentences now that $\phi$ is part of it. But still, I don't have any idea how to prove that $T_0$ is deductively closed.
 A: So, as @Taroccoesbrocco has pointed out, $\mathcal{L}_0$ refers to the set of all formulas in the language $\mathcal{L}_0$, so it is not just any subset of $\mathcal{L}$:
$$
\mathcal{L}_0\vDash\alpha\Leftrightarrow\alpha\in\mathcal{L}_0
$$
Now on to the proof that $T_0:=T\cap\mathcal{L}_0$ is a theory.
It is clear that $T_0$ only contains sentences, because $T$ only contains sentences and thus the same is true for $T_0\subseteq T$. What is left to prove is that $T_0$ is deductively closed:
$$
\begin{align*}
T_0\vDash\alpha&\Rightarrow T\vDash\alpha\ \mathrm{and}\ \mathcal{L}_0\vDash\alpha\text{  (follows from monotonicity of }\vDash\text{)}\\
&\Rightarrow \alpha\in T\ \mathrm{and}\ \alpha\in\mathcal{L}_0\\
&\Rightarrow \alpha\in T_0
\end{align*}
$$
$T_0$ is a subset both of $T$ and of $\mathcal{L}_0$. The monotonicity of $\vDash$ states that if $X\vDash\alpha$ for some set of formulas $X$, then $X'\vDash\alpha$ for $X'\supseteq X$.
The other direction, $\alpha\in T_0\Rightarrow T_0\vDash\alpha$, is obvious.
