X compact iff any collection of closed set closed by finite intersection have a nonempty intersection I've got a problem to show something. Actually, we consider $X$ a topological space , and we want to show :
$X$ compact $\iff \forall (F_i)_{i \in I} $ collection of closed set which is closed by finite intersection : $\bigcap_{i \in I} F_i \neq \emptyset$
It's okay for the direction $\implies$. But for the other direction, I'm stuck. Actually, this is what I did :
Let $(O_i)_{i \in I} \in (\mathcal{O}_X)^I$ such that : $X = \bigcup_{i \in I} O_i$. If there is $i_0 \in I$ such that $U_{i_0} = X$, then $X = U_{i_0}$ and it's okay. Otherwise, we have :
$X = (\bigcap_{i \in I} U_i{}^c)^c$, and we consider the collection : $(U_i^{c})_{i \in I}$ which is a collection of closed subset. Then, I tried different things, I mainly try to introduce a set as : $\mathcal{A} = \{ J \subset I \; | \; J$ finite and $ (U_j^{c})_{j \in J}$ closed by finite intersection $\}$ or other set like this in order to find a $J \subset I$, $J$ a finite set, which verify : $\bigcup_{j \in J} U_j = X$. But I didn't find.
If someone could help me, it will be very appreciated ! 
Thank you !
 A: For the reverse direction I would recommend arguing by contrapositive.
Claim: If $X$ is not compact, then there exists a family of closed sets $\{U_{\alpha}\}_{\alpha \in I}$ that is closed under finite intersections, such that $\bigcap\limits_{\alpha\in I} U_{\alpha} = \varnothing.$
Proof: Suppose $X$ is not compact. Then, there exists an open cover $\{V_{\alpha}\}_{\alpha \in I}$ for $X$ which does not admit a finite subcover. We will assume without loss of generality that for all $\alpha \in I$, $V_{\alpha} \neq \varnothing$. As such, we may for all $\alpha \in I$ define $V_{\alpha} = U_{\alpha}^{C}$ so that $\mathscr{G}=\{U_{\alpha}\}_{\alpha \in I} \subseteq \mathscr{P}(X)$ is a family of nonempty, closed sets. Fix a finite subset $J \subset I$. Since $\{V_{\alpha}\}_{\alpha \in I}$ does not admit a finite subcover, 
$$\varnothing \neq \bigcup\limits_{\alpha \in J} V_{\alpha} \subset X \implies \varnothing \neq \bigg(\bigcup\limits_{\alpha \in J} V_{\alpha}\bigg)^{C} \subset X. $$ 
But from our definition of $\{U_{\alpha}\}_{\alpha \in I}$ and de Morgan's Law, it follows that
$$\bigcap\limits_{\alpha \in J}U_{\alpha}=\bigcap\limits_{\alpha\in J}V_{\alpha}^{C}=\bigg(\bigcup\limits_{\alpha \in J} V_{\alpha}\bigg)^{C} \neq \varnothing.$$
Since $J \subset I$ was an arbitrary finite set, we conclude that $\{U_{\alpha}\}_{\alpha \in I}$ has the finite intersection property. Now constructing the set 
$$\mathscr{A}:=\{J \in \mathscr{P}(I):\textrm{J finite}\}$$ and letting $I'$ be an indexing set for $\mathscr{A}$ so that $\mathscr{A}=\{J_{\beta}\}_{\beta \in I'}$, we in turn construct the family of sets $\mathscr{H}=\{W_{\beta}\}_{\beta \in I'}$ such that for each $\beta \in I'$, $W_{\beta}=\bigcap\limits_{\alpha \in J_{\beta}}U_{\alpha}$. Further defining $\mathscr{I}=I \cup I'$, and $\mathscr{F}=\mathscr{G} \cup \mathscr{H}$, we see that $\mathscr{F} =\{K_{\alpha}\}_{\alpha \in \mathscr{I}}\subseteq \mathscr{P}(X)$ is a family of closed sets which is closed under finite intersections, with $$\bigcap\limits_{\alpha \in \mathscr{I}}K_{\alpha}=\bigcap\limits_{\alpha \in I}U_{\alpha}.$$
However, since the $\{V_{\alpha}\}_{\alpha \in I}$ are an open cover for $X$ by assumption, we may again invoke the definition of $\{U_{\alpha}\}_{\alpha \in I}$ to show that
$$\bigcap\limits_{\alpha \in \mathscr{I}}K_{\alpha}=\bigcap_{\alpha \in I}U_{\alpha}=\bigcap_{\alpha \in I}V_{\alpha}^{C}=\bigg(\bigcup\limits_{\alpha \in I} V_{\alpha}\bigg)^{C} =X^{C}=\varnothing,$$
which is what we promised to show.
Regards!
-Dan
