[proofplan]
We prove the equivalence by passing between closed sets and their open complements. If $X$ is compact and a closed family has empty total intersection, then the complementary open sets cover $X$, so compactness gives finitely many complements that already cover $X$; the corresponding finite intersection of closed sets is empty, contradicting the finite intersection property. Conversely, if every closed family with the finite intersection property has nonempty total intersection, then an [open cover](/page/Open%20Cover) without a [finite subcover](/page/Finite%20Subcover) produces closed complements with the finite intersection property; their nonempty total intersection gives a point missed by the cover, which is impossible.
[/proofplan]
[step:Turn an empty closed intersection into an open cover]
Assume first that $X$ is compact. Let $\mathcal F$ be a family of closed subsets of $X$ with the finite intersection property. Suppose, toward a contradiction, that
\begin{align*}
\bigcap_{F \in \mathcal F} F = \varnothing.
\end{align*}
For each $F \in \mathcal F$, define the open complement
\begin{align*}
U_F := X \setminus F.
\end{align*}
Since $F$ is closed in $(X,\tau)$, we have $U_F \in \tau$. We claim that the family
\begin{align*}
\mathcal U := \{U_F : F \in \mathcal F\}
\end{align*}
is an open cover of $X$.
Indeed, let $x \in X$. Since the total intersection of the members of $\mathcal F$ is empty, $x$ does not belong to every $F \in \mathcal F$. Hence there exists $F_x \in \mathcal F$ such that $x \notin F_x$. Therefore $x \in X \setminus F_x = U_{F_x}$, so
\begin{align*}
X \subset \bigcup_{F \in \mathcal F} U_F.
\end{align*}
[guided]
Assume that $X$ is compact, and let $\mathcal F$ be a family of closed subsets of $X$ with the finite intersection property. We argue by contradiction and suppose that the total intersection is empty:
\begin{align*}
\bigcap_{F \in \mathcal F} F = \varnothing.
\end{align*}
The natural way to use compactness is through open covers, so we convert the closed family into an open family by taking complements. For each $F \in \mathcal F$, define
\begin{align*}
U_F := X \setminus F.
\end{align*}
This set is open in the topology $\tau$ because $F$ is closed in $X$.
Now we verify that these open complements cover $X$. Let $x \in X$ be arbitrary. The equality
\begin{align*}
\bigcap_{F \in \mathcal F} F = \varnothing
\end{align*}
means that no point of $X$ lies in every member of $\mathcal F$. Therefore this particular point $x$ fails to belong to at least one member of the family: there exists $F_x \in \mathcal F$ such that $x \notin F_x$. By the definition of $U_{F_x}$, this is exactly the statement that $x \in U_{F_x}$. Since $x \in X$ was arbitrary, the family
\begin{align*}
\mathcal U := \{U_F : F \in \mathcal F\}
\end{align*}
satisfies
\begin{align*}
X \subset \bigcup_{F \in \mathcal F} U_F.
\end{align*}
Thus $\mathcal U$ is an open cover of $X$.
[/guided]
[/step]
[step:Use compactness to contradict the finite intersection property]
Since $\mathcal U$ is an open cover of the [compact space](/page/Compact%20Space) $X$, there exists a finite subfamily $\mathcal E \subset \mathcal F$ such that
\begin{align*}
X \subset \bigcup_{F \in \mathcal E} U_F.
\end{align*}
Using $U_F = X \setminus F$, this says
\begin{align*}
X \subset \bigcup_{F \in \mathcal E} (X \setminus F).
\end{align*}
Taking complements in $X$, we obtain
\begin{align*}
X \setminus \bigcup_{F \in \mathcal E} (X \setminus F) = \varnothing.
\end{align*}
By the elementary complement identity
\begin{align*}
X \setminus \bigcup_{F \in \mathcal E} (X \setminus F) = \bigcap_{F \in \mathcal E} F,
\end{align*}
it follows that
\begin{align*}
\bigcap_{F \in \mathcal E} F = \varnothing.
\end{align*}
This contradicts the finite intersection property of $\mathcal F$. Therefore
\begin{align*}
\bigcap_{F \in \mathcal F} F \ne \varnothing.
\end{align*}
This proves the forward implication.
[/step]
[step:Convert an open cover without a finite subcover into closed complements]
Conversely, assume that every family of closed subsets of $X$ with the finite intersection property has nonempty total intersection. Let $\mathcal U$ be an open cover of $X$. We prove that $\mathcal U$ has a finite subcover.
Suppose, toward a contradiction, that no finite subfamily of $\mathcal U$ covers $X$. For each $U \in \mathcal U$, define the closed complement
\begin{align*}
F_U := X \setminus U.
\end{align*}
Since $U \in \tau$, the set $F_U$ is closed in $X$. Let
\begin{align*}
\mathcal F := \{F_U : U \in \mathcal U\}.
\end{align*}
We show that $\mathcal F$ has the finite intersection property. Let $\mathcal V \subset \mathcal U$ be finite. Because $\mathcal V$ is not a finite subcover of $X$, there exists $x_{\mathcal V} \in X$ such that
\begin{align*}
x_{\mathcal V} \notin \bigcup_{U \in \mathcal V} U.
\end{align*}
Equivalently, $x_{\mathcal V} \in X \setminus U = F_U$ for every $U \in \mathcal V$. Hence
\begin{align*}
x_{\mathcal V} \in \bigcap_{U \in \mathcal V} F_U.
\end{align*}
Thus every finite subfamily of $\mathcal F$ has nonempty intersection, so $\mathcal F$ has the finite intersection property.
[guided]
Now assume the finite-intersection-property condition for closed families, and let $\mathcal U$ be an open cover of $X$. To prove compactness, we must show that $\mathcal U$ has a finite subcover.
Suppose instead that no finite subfamily of $\mathcal U$ covers $X$. We again pass to complements, but now in the opposite direction: for each $U \in \mathcal U$, define
\begin{align*}
F_U := X \setminus U.
\end{align*}
Because $U$ is open in $X$, the set $F_U$ is closed in $X$. Let
\begin{align*}
\mathcal F := \{F_U : U \in \mathcal U\}.
\end{align*}
We need to verify the finite intersection property for $\mathcal F$. Take any finite subfamily of $\mathcal F$. Such a subfamily is obtained from finitely many open sets in the cover; write those open sets as a finite family $\mathcal V \subset \mathcal U$. Since we assumed that no finite subfamily of $\mathcal U$ covers $X$, the family $\mathcal V$ does not cover $X$. Therefore there exists a point $x_{\mathcal V} \in X$ outside all sets in $\mathcal V$, meaning
\begin{align*}
x_{\mathcal V} \notin \bigcup_{U \in \mathcal V} U.
\end{align*}
This is equivalent to saying that, for every $U \in \mathcal V$,
\begin{align*}
x_{\mathcal V} \in X \setminus U = F_U.
\end{align*}
Consequently,
\begin{align*}
x_{\mathcal V} \in \bigcap_{U \in \mathcal V} F_U.
\end{align*}
So the intersection of the corresponding finite subfamily of closed complements is nonempty. Since the finite family was arbitrary, $\mathcal F$ has the finite intersection property.
[/guided]
[/step]
[step:Derive the contradiction and obtain a finite subcover]
By the assumed finite-intersection-property condition, the closed family $\mathcal F$ satisfies
\begin{align*}
\bigcap_{F \in \mathcal F} F \ne \varnothing.
\end{align*}
Choose
\begin{align*}
x \in \bigcap_{F \in \mathcal F} F.
\end{align*}
Then for every $U \in \mathcal U$, the point $x$ belongs to $F_U = X \setminus U$, so $x \notin U$. Hence
\begin{align*}
x \notin \bigcup_{U \in \mathcal U} U.
\end{align*}
This contradicts the assumption that $\mathcal U$ is an open cover of $X$. Therefore every open cover of $X$ has a finite subcover, and $X$ is compact.
[/step]
