Prove $X$ is compact if every family of closed sets with the finite intersection property have non-empty intersection 
Let $X$ a metric space, and $\mathcal M$ a family of closed subsets of $X$ with the finite intersection property, i.e. with the property that every finite intersection of elements of $\mathcal M$ is non-empty.
Show that $X$ is compact if and only if every family of closed sets with the finite intersection property have non-empty intersection.

Let define $\mathcal M:=\{A_\alpha\subset  X:A_\alpha=\overline A_\alpha\land \alpha\in I\}$ with the properties defined above, i.e.
$$\bigcap \mathcal M\neq\emptyset\iff\bigcap_{\alpha\in J} A_\alpha\neq\emptyset,\;\forall J\subseteq I:|J|\in\Bbb N$$
(The right implication is obvious.)
My strategy for this proof is to show that $X$ is complete and totally bounded, what is equivalent to be compact.

Proof:
$X$ is complete: for every Cauchy sequence $(x_n)$ in $X$ start choosing $X$ as a closed neighborhood for $x_0$. Because $(x_n)$ is Cauchy for every $\epsilon=1/k$ exists some $N\in\Bbb N$ such that $d(x_n,x_m)<1/k$ whenever $n,m\ge N$.
Then for every Cauchy sequence on $X$ we can define the family of sets
$$\mathcal B:=\{B_k:B_k=\overline{\Bbb B}(x_N,1/k), k\in\Bbb N_{>0}\land B_0=X\}$$
what is a family of nested closed sets with the property of non-empty finite intersection. Then if $\bigcap\mathcal B\neq\emptyset$ then $X$ is complete, i.e every Cauchy sequence converges in $X$.
$X$ is totally bounded: if $X$ is not totally bounded then exists some $\epsilon>0$ such that doesnt exists a finite open cover of $X$ composed of open balls with centers in $X$. Then for some $\epsilon>0$ the family
$$C_\epsilon :=\{\Bbb B(x,\epsilon):x\in X\}$$
doesn't contain a finite subcover. Now define the family composed by the complements of $C_\epsilon$:
$$C_\epsilon^*:=\{(\Bbb B(x,\epsilon))^\complement:x\in X\}$$
then observe that the family $C_\epsilon^*$ have the finite intersection property because in other case $C_\epsilon$ would have a finite subcover.
But we have that
$$\bigcap C_\epsilon^*=\bigcap (\Bbb B(x,\epsilon))^\complement=\left(\bigcup \Bbb B(x,\epsilon)\right)^\complement=X^\complement=\emptyset$$
what is a contradiction about the assumptions on $X$, then $X$ is totally bounded.
Because $X$ is totally bounded and complete then is compact.$\Box$


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*Can you check this proof please and comment any dubious or wrong step?


*If you know some alternative proof easier to this approach it will be very interesting to see it. Thank you.
 A: 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 $\bigcap_{i\in I}C_i\neq\emptyset$.
