If $x$ is a set then $\cap x$ is a set Consider the following problem: 

(*) If $A$ is a non-empty set, then $\cap A$ is a set.

But I don't think we need the hypothesis "$A$ is non-empty". I can simply write $$\{x \in A \mid \forall y (y \in A ) \rightarrow (x \in y)\}$$ This is a set by the separation axiom and it is the intersection set $\cap A$, isn't it?

What did I do wrong?
Let $A=\{\{a\}\}$, say. Then $\bigcap A=\{a\}$ but $a\notin A$, so your proposed
definition gives $\emptyset$.
Axiom: $\cup A=\{x:\exists y\in A\,(x\in y)\}$ exists. So let $B=\{x\in \cup A: \forall y\in A\,(x\in y)\},$ which exists by Comprehension (Separation). If $A=\emptyset$ then $\cup A=\emptyset$ so $B=\emptyset.$ If $A\ne \emptyset$ then $B=\{x:\forall y\in A\,(x\in y)\}=\cap A$.