There's a row of models, namely $M_0 \subseteq M_1 \subseteq M_2 \subseteq \dotsb$ and there's a theory $T$ such: $\phi \in T \rightarrow \phi = \forall x\exists y\,\psi$ with $\psi$ quantor-free. Prove that:
\begin{equation} \left[\left(\forall i \in \mathbb N\right)\left(M_i \vDash T\right)\right] \rightarrow \bigcup M_i \vDash T \end{equation}
My idea was that $m$ from $\forall m\exists m'\,\psi$ is in a certain $M_i$ and in that $M_i$ there exists an $m'$ (because $M_i$ is a model). Is this correct?
Your notation is a little confusing, but I think you have the right idea:
Note that $\bigcup M_i\models \forall x\exists y\;\psi (x,y) \Longleftrightarrow$ for every $m\in \bigcup M_i$, $\bigcup M_i\models \exists y\;\psi (m,y)$.
Let $m\in \bigcup M_i$, then $m\in M_i$ for some $i$. As $M_i\models T$, there is $m'\in M_i$ such that $M_i\models \psi (m,m')$, so $\bigcup M_i\models \psi (m,m')$ and hence $\bigcup M_i\models \exists y\;\psi (m,y)$.