How to prove that for $A\cap B\neq\varnothing$, $(\bigcap A)\cap(\bigcap B)\subseteq\bigcap(A\cap B)$? $A$ and $B$ are non empty sets with non empty intersection.
Prove that
$(\bigcap A)\cap(\bigcap B) \subseteq \bigcap (A\cap B).$
The definition of intersection of a set is something like this, if $M$ is a nonempty set whose elements are themselves sets, then $x$ is an element of the intersection of $M$ if and only if for every element $A$ of $M$, $x$ is an element of $A$.

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.
