Claim: A topological space $\,X\,$ is compact iff it has the Finite Intersection Property (=FIP):
Proof: (1) Suppose $\,X\,$ is compact and let $\,\{V_i\}\,$ be a family of closed subsets s.t. $\,\displaystyle{\bigcap_{i}V_i=\emptyset}\,$. Putting now $\,A_i:=X-V_i\,$ , we get that $\,\{A_i\}\,$ is a family of open subsets , and
$$\bigcap_{i}V_i=\emptyset\Longrightarrow \;X=X-\emptyset=X-\left(\bigcap_iV_i\right)=\bigcup_i\left(X-V_i\right)=\bigcup_iA_i\Longrightarrow\;\{A_i\}$$
is an open cover of $\,X\,$ and thus there exists a finite subcover of it:
$$X=\bigcup_{i\in I\,,\,|I|<\aleph_0}A_i=\bigcup_{i\in I\,,\,|I|<\aleph_0}(X-V_i)=X-\left(\bigcap_{i\in I\,,\,|I|<\aleph_0}V_i\right)\Longrightarrow \bigcap_{i\in I\,,\,|I|<\aleph_0}V_i=\emptyset\Longrightarrow$$
The family $\,\{V_i\}\,$ has the FIP.
(2) Suppose now that every family of closed subsets of $\,X\,$ hast the FIP, and let $\,\{A_i\}\,$ be an open cover of it. Put $\,U_i:=X-A_i\,$ , so $\,U_i\, $ is closed for every $\,i\,$:
$$\bigcap_iU_i=\bigcap_i(X-A_i)=X-\bigcup_i A_i=X-X=\emptyset$$
By assumption, there exists a finite set $\,J\,$ s.t. $\,\displaystyle{\bigcap_{i\in J}U_i=\emptyset}\,$ , but then
$$X=X-\emptyset=X-\bigcap_{i\in J}U_i=\bigcup_{i\in J}X-U_i)=\bigcup_{i\in J}A_i\Longrightarrow$$
$\,\{A_i\}_{i\in J}$ is a finite subcover for $\,X\,$ and thus it is compact....QED.
Please be sure you can follow the above and justify all steps. Check where we used Morgan Rules, for example, and note that we used the contrapositive of the FIP's definition...
This theorem holds also for topological spaces that are not metric, so using Cauchy sequences might not be the most suitable approach.
Let $X$ be a topological space and assume that the intersection of any family of closed subsets having the finite intersection property is non-empty. Let $\{O_i\}_{i\in I}$ be some open cover of $X$ and assume that it has no finite subcover. Hence for each finite $F\subseteq I$ there is some $x\in X$ such that $x\notin \bigcup_{i\in F}O_i$. Equivalently, for each finite $F\subseteq I$ there is some $x\in\bigcap_{i\in I}(X\setminus O_i)$. Hence the family $\{X\setminus O_i\}_{i\in I}$ is a family of closed subsets with the finite intersection property. It follows that there is some $x\in\bigcap_{i\in I}(X\setminus O_i)$, i.e., $x\notin\bigcup_{i\in I}O_i$, contradicting that $\{O_i\}_{i\in I}$ is a cover of $X$. We conclude that $\{O_i\}_{i\in I}$ must have a finite subcover, so $X$ is compact.
Conversely, assume that $X$ is compact and let $\{C_i\}_{i\in I}$ be some family of closed subsets with the finite intersection property. Assume that $\bigcap_{i\in I}C_i$ is empty. Then $\{X\setminus C_i\}_{i\in I}$ is an open cover of $X$. By compactness, it has a finite subcover, so $\bigcup_{i\in F}X\setminus C_i=X$ for some finite $F\subseteq I$. But then $\bigcap_{i\in F}C_i=\emptyset$, contradicting that $\{C_i\}_{i\in I}$ has the finite intersection property. We conclude that $\bigcap_{i\in I}C_i\neq\emptyset$.
Best Answer
Although your proof goes well, but it needs more explanation. Here is an attempt.
Suppose that $X$ is compact and let $\{F_\alpha:\alpha\in I\}$ be a family of closed subsets of $X$ with $FIP$. On contrary, suppose that $\cap_{\alpha\in I}F_\alpha=\emptyset$. Then $\cup_{\alpha\in I}F_\alpha^c=X$. As $X$ is compact, so there is a finite subset $J$ of $I$ such that $\cup_{\alpha\in J}F_\alpha^c=X$ and consequently $\cap_{\alpha\in J}F_\alpha^c=\emptyset$, which is a contradiction to $FIP$.
Conversely, let $\{U_\alpha:\alpha\in I\}$ be an open cover of $X$. Now put $F_\alpha=U_\alpha^c$. If $X$ is not compact then the family $\{F_\alpha:\alpha\in I\}$, of closed subsets of $X$, has $FIP$ with $\cap_{\alpha\in I}F_\alpha=\emptyset$, which contradicts our hypothesis.