[proofplan]
We prove the two constructions directly from the definitions. For a closure operator, the whole set $E$ is closed, and arbitrary intersections of closed sets are fixed by using monotonicity against each member of the intersection. Conversely, a Moore family gives a closure by intersecting all Moore-family members containing a set; the Moore property guarantees that this intersection exists inside the family, and the closure axioms follow from elementary containment relations among the indexing families. Finally, the fixed points of this constructed operator are exactly the Moore-family members.
[/proofplan]
[step:Show that fixed points of a closure operator form a Moore family]
Let $c:2^E\to 2^E$ be a closure operator, and define
\begin{align*}
\mathcal C=\{F\subseteq E:c(F)=F\}.
\end{align*}
First, $E\in\mathcal C$. Since $c(E)\subseteq E$ because $c(E)\in 2^E$, and $E\subseteq c(E)$ by extensivity, we have $c(E)=E$.
Now let $(F_i)_{i\in I}$ be an indexed family of elements of $\mathcal C$, where $I$ is any index set. Define
\begin{align*}
F=\bigcap_{i\in I}F_i,
\end{align*}
with the convention that the empty intersection is $E$. If $I=\varnothing$, then $F=E\in\mathcal C$. Suppose $I\neq\varnothing$. For every $i\in I$, we have $F\subseteq F_i$, so monotonicity gives
\begin{align*}
c(F)\subseteq c(F_i)=F_i.
\end{align*}
Thus $c(F)\subseteq \bigcap_{i\in I}F_i=F$. Extensivity gives $F\subseteq c(F)$, hence $c(F)=F$. Therefore $F\in\mathcal C$, so $\mathcal C$ is closed under arbitrary intersections and contains $E$. Hence $\mathcal C$ is a Moore family.
[guided]
Let us check the Moore-family axioms one at a time. We define
\begin{align*}
\mathcal C=\{F\subseteq E:c(F)=F\}.
\end{align*}
This is the family of fixed points of the closure operator. The first Moore-family requirement is that $E$ itself belongs to $\mathcal C$. Since $c$ maps subsets of $E$ to subsets of $E$, we have $c(E)\subseteq E$. On the other hand, extensivity applied to $A=E$ gives
\begin{align*}
E\subseteq c(E).
\end{align*}
The two inclusions imply $c(E)=E$, so $E\in\mathcal C$.
Now we prove closure under arbitrary intersections. Let $(F_i)_{i\in I}$ be any indexed family of sets in $\mathcal C$, and define
\begin{align*}
F=\bigcap_{i\in I}F_i.
\end{align*}
If $I=\varnothing$, then by convention $F=E$, and we already proved that $E\in\mathcal C$. Assume therefore that $I\neq\varnothing$. For each index $i\in I$, the definition of intersection gives $F\subseteq F_i$. Since $c$ is monotone, applying $c$ to this inclusion gives
\begin{align*}
c(F)\subseteq c(F_i).
\end{align*}
Because $F_i\in\mathcal C$, each $F_i$ is fixed by $c$, so $c(F_i)=F_i$. Hence
\begin{align*}
c(F)\subseteq F_i
\end{align*}
for every $i\in I$. Therefore $c(F)$ is contained in the intersection of all the $F_i$:
\begin{align*}
c(F)\subseteq \bigcap_{i\in I}F_i=F.
\end{align*}
The reverse inclusion is exactly extensivity applied to $F$:
\begin{align*}
F\subseteq c(F).
\end{align*}
Combining the two inclusions gives $c(F)=F$, so $F\in\mathcal C$. Thus arbitrary intersections of members of $\mathcal C$ remain in $\mathcal C$, and $\mathcal C$ is a Moore family.
[/guided]
[/step]
[step:Construct the closure operator associated to a Moore family]
Let $\mathcal M\subseteq 2^E$ be a Moore family. For each $A\subseteq E$, define the indexing family
\begin{align*}
\mathcal M_A=\{M\in\mathcal M:A\subseteq M\}.
\end{align*}
Since $E\in\mathcal M$ and $A\subseteq E$, we have $E\in\mathcal M_A$, so $\mathcal M_A$ is nonempty. Define
\begin{align*}
c_{\mathcal M}:2^E\to 2^E
\end{align*}
by
\begin{align*}
c_{\mathcal M}(A)=\bigcap_{M\in\mathcal M_A}M.
\end{align*}
Because $\mathcal M$ is closed under arbitrary intersections, $c_{\mathcal M}(A)\in\mathcal M$ for every $A\subseteq E$.
[/step]
[step:Verify extensivity and monotonicity for the constructed operator]
Let $A\subseteq E$. Every member $M\in\mathcal M_A$ contains $A$, so
\begin{align*}
A\subseteq \bigcap_{M\in\mathcal M_A}M=c_{\mathcal M}(A).
\end{align*}
Thus $c_{\mathcal M}$ is extensive.
Now let $A,B\subseteq E$ with $A\subseteq B$. If $M\in\mathcal M_B$, then $B\subseteq M$, and therefore $A\subseteq M$, so $M\in\mathcal M_A$. Hence
\begin{align*}
\mathcal M_B\subseteq \mathcal M_A.
\end{align*}
Taking intersections reverses this inclusion of indexing families, so
\begin{align*}
\bigcap_{M\in\mathcal M_A}M\subseteq \bigcap_{M\in\mathcal M_B}M.
\end{align*}
Therefore $c_{\mathcal M}(A)\subseteq c_{\mathcal M}(B)$, and $c_{\mathcal M}$ is monotone.
[/step]
[step:Use membership in the Moore family to prove idempotence]
Let $A\subseteq E$. From the construction, $c_{\mathcal M}(A)\in\mathcal M$. Extensivity gives $A\subseteq c_{\mathcal M}(A)$, so $c_{\mathcal M}(A)\in\mathcal M_A$. Therefore the intersection defining $c_{\mathcal M}(A)$ is contained in this particular member:
\begin{align*}
c_{\mathcal M}(A)\subseteq c_{\mathcal M}(A).
\end{align*}
More importantly, since $c_{\mathcal M}(A)\in\mathcal M$ and $c_{\mathcal M}(A)\subseteq c_{\mathcal M}(A)$, the set $c_{\mathcal M}(A)$ belongs to $\mathcal M_{c_{\mathcal M}(A)}$. Hence
\begin{align*}
c_{\mathcal M}(c_{\mathcal M}(A))\subseteq c_{\mathcal M}(A).
\end{align*}
The reverse inclusion follows from extensivity applied to $c_{\mathcal M}(A)$:
\begin{align*}
c_{\mathcal M}(A)\subseteq c_{\mathcal M}(c_{\mathcal M}(A)).
\end{align*}
Thus
\begin{align*}
c_{\mathcal M}(c_{\mathcal M}(A))=c_{\mathcal M}(A).
\end{align*}
So $c_{\mathcal M}$ is idempotent.
[/step]
[step:Identify the closed sets of the constructed closure operator]
We prove that the fixed points of $c_{\mathcal M}$ are exactly the elements of $\mathcal M$.
First let $F\in\mathcal M$. Since $F\subseteq F$, we have $F\in\mathcal M_F$. Thus
\begin{align*}
c_{\mathcal M}(F)=\bigcap_{M\in\mathcal M_F}M\subseteq F.
\end{align*}
Extensivity gives $F\subseteq c_{\mathcal M}(F)$, so $c_{\mathcal M}(F)=F$. Hence every element of $\mathcal M$ is closed.
Conversely, let $F\subseteq E$ satisfy $c_{\mathcal M}(F)=F$. By construction, $c_{\mathcal M}(F)$ is an arbitrary intersection of elements of $\mathcal M$, so the Moore property gives $c_{\mathcal M}(F)\in\mathcal M$. Since $F=c_{\mathcal M}(F)$, it follows that $F\in\mathcal M$.
Therefore the closed sets of $c_{\mathcal M}$ are precisely the elements of $\mathcal M$. Combining this with the previous steps proves the equivalence between closure operators on $E$ and Moore families on $E$.
[/step]
