For theorems like these, as Asaf wrote, expanding definitions and simplifying is the way to go. However, I do these kind of things more 'calculationally' using the rules of predicate logic.
In this case, we can easily calculate the elements $\;x\;$ of the left hand side: \begin{align} & x \in \bigcap A \;\cap\; \bigcap B \\ \equiv & \qquad\text{"definition of $\;\cap\;$; definition of $\;\bigcap\;$, twice"} \\ & \langle \forall V : V \in A : x \in V \rangle \;\land\; \langle \forall V : V \in B : x \in V \rangle \\ \equiv & \qquad\text{"logic: merge ranges of $\;\forall\;$ statements -- to simplify"} \\ & \langle \forall V : V \in A \lor V \in B : x \in V \rangle \\ \end{align}
And similarly for the right hand side: \begin{align} & x \in \bigcap (A \cap B) \\ \equiv & \qquad\text{"definition of $\;\bigcap\;$; definition of $\;\cap\;$"} \\ & \langle \forall V : V \in A \cap B : x \in V \rangle \\ \equiv & \qquad\text{"definition of $\;\cup\;$"} \\ & \langle \forall V : V \in A \land V \in B : x \in V \rangle \\ \end{align}
These two results look promisingly similar. We see that latter range implies the former, and predicate logic tells us that $ \langle \forall z : P(z) : R(z) \rangle \;\Rightarrow\; \langle \forall z : Q(z) : R(z) \rangle $ if $\;Q(z) \Rightarrow P(z)\;$ for all $\;z\;$. In our specific case, that means \begin{align} & \langle \forall V : V \in A \lor V \in B : x \in V \rangle \\ \Rightarrow & \qquad \text{"using the above rule, with $\;P \land Q \;\Rightarrow\; P \lor Q\;$"} \\ & \langle \forall V : V \in A \land V \in B : x \in V \rangle \\ \end{align}
Putting this all together, with the definition of $\;\subseteq\;$, tells us that $ \bigcap A \;\cap\; \bigcap B \;\subseteq\; \bigcap (A \cap B) $ which is what we set out to prove.