Prove that the arbitrary intersection indexed over the power set is the empty set and that the arbitrary union indexed over the power set is the entire set itself.
$a) \bigcap_{A\in P(X)} A = \varnothing $
$b) \bigcup_{A\in P(X)} A = X $
Where $P(X)$ is the power set of $X$
Proof:-
a) Let $x \in \bigcap_{A \in P(X)}A$
Then $\forall A\subseteq X\quad (or\quad \forall A \in P(X)$)
$x \in A $
Therefore $\bigcap_{A \in P(X)}A \subseteq A \quad (\forall A\in P(X)$)
Now $\varnothing \in P(X)$
Thus $\bigcap_{A \in P(X)}A \subseteq \varnothing$
We know that the empty set is a subset of every set, in particular:
$\varnothing \subseteq \bigcap_{A \in P(X)}A$
Therefore we've demonstrated set inclusion in both directions and by the definition of set equality, it follows that,
$\bigcap_{A \in P(X)}A = \varnothing$.
b) Let $x \in \bigcup_{A \in P(X)}A$
Then
$\exists A \in P(X) $ such that $x\in A$
and since $A \subseteq X $ it follows that $x \in A$
Therefore $\bigcup_{A \in P(X)}A \subseteq X$
Conversely, let $x \in X$
Then $\exists A \in P(X)$ such that $ x \in A $
And, so $x \in\bigcup_{A \in P(X)}A$
Therefore we've demonstrated set inclusion in both directions and by the definition of set equality, it follows that,
$\bigcup_{A \in P(X)}A=X$.
Quod Erat Demonstrandum.
a) You stated correctly that an element $x$ of the intersection must be an element of every $A\subseteq X$. For $A$ we can take $\varnothing$ so the assumption that $x$ is an element of the intersection leads directly to the conclusion that $x\in\varnothing$. This is absurd so you are allowed to reject the assumption. This for every $x\in X$ and the final conclusion is then that the intersection is empty.
b)
I suspect a typo here and that you meant to end this with "...that $x\in X$"
(You stated correctly that - if $x$ is an element of the union - we have $x\in A$ for some $A\subseteq X$. From this you can conlude directly that $x\in X$.)