How would you go about proving this with a formal proof? 
For a set $\mathscr{F}$, define
$$\bigcup!\mathscr{F}=\{x\mid\exists!A\in\mathscr{F}\, x\in A\}$$
(a) If $\mathscr{F}=\big\{\{1,2,3\},\{2,3,4\},\{3,4,5\}\big\}$, what is $\bigcup!\mathscr{F}$?
(b) Suppose $\mathscr{F}$ is a set and $\bigcup!\mathscr{F}\cap\bigcap\mathscr{F}\ne\varnothing$. Show that $\mathscr{F}$ has exactly one element.

I do not know how to write a formal proof to these, please could someone teach me how to write these types of formal proofs? I am pretty sure it has to do with existence and uniqueness.
 A: For (b). Assume $[\cup !\mathscr F ]\cap [\cap \mathscr F]\ne \emptyset.$
If $f,g\in\mathscr F$ with $f\ne g$ then $\cup !\mathscr F\subseteq (f\setminus g)\cup (g\setminus f)$ and $\cap\mathscr F)\subseteq f\cap g.$ But for any $f,g$ we have $[(f\setminus g)\cup (g\setminus f)]\,\cap \, [f\cap g]=\emptyset.$ Therefore $\mathscr F$ has at most 1 member.
If $\mathscr F=\emptyset$ then $[\cup !\mathscr F ]\cap [\cap \mathscr F]\subseteq\cup !\mathscr F=\emptyset.$ Therefore $\mathscr F$ has at least 1 member.
A: $(a)$ is $\{1,5\}$, the full union is $\{1,2,3,4,5\}$ but $2,3$ and $4$ occur in two sets in $\mathcal{F}$, so are not in $\bigcup !\mathcal{F}$.
$(b)$: Let $x \in \bigcup !\mathcal{F}$ and $x \in \bigcap \mathcal{F}$.
Suppose that $F_1, F_2 \in \mathcal{F}$ exist with $F_1 \neq F_2$.
Then $x \in F_1 \cap F_2$ as $x$ lies in all members of $\mathcal{F}$ by assumption. But then $x$ cannot be in $\bigcup ! \mathcal{F}$ by definition as it is not in a unique member of $\mathcal{F}$, but in at least two. This contradiction shows that $\mathcal{F}$ cannot have two distinct elements and as no elements is ruled out too (an empty union is always empty, also with $!$) $\mathcal{F}$ has a single member, as required.
