$T$ is a theory and $\phi$ is a sentence with $T \models \phi$. I read notes with a quote like this:
By Compactness Theorem, a finite subset $T_0 \subseteq T$ has $T_0 \models \phi$.
I thought the Compactness Theorem was something like "a theory has a model iff every subset of the theory has a model". That is $M \models T \implies M \models T_0$. (I believe it follows from Completeness of FOL and proofs being finite). So how do we show the claim with compactness? I think it has something to do with $\phi$ being a sentence. If we replaced $\phi$ with an infinite theory $T'$ then we cannot claim $T_0 \models T'$.
Your statement of compactness is not quite right (your statement is true, just not very strong, since $T$ is a subset of itself!). A correct statement is: Let $T$ be a first-order theory. Then $T$ has a model if and only if every finite subset of $T$ has a model.
Now to your actual question. Since $T \models \phi$, the theory $T \cup \{\neg\phi\}$ has no models. Thus, by compactness, there is a finite subset $\Delta \subseteq T \cup \{\neg\phi\}$ that has no models. Let $T_0 \subseteq T$ be the finitely many sentences from $T$ that appear in $\Delta$. Then $T_0 \cup \{\neg\phi\}$ has no models, so every model of $T_0$ is not a model of $\neg\phi$, and hence $T_0 \models \phi$.