The more general fact is true. If $(C_\alpha)_{\alpha\in I}$ is a collection of closed subsets with finite intersection property, then all these subsets have a common point.
Assume $\cap_{\alpha\in I}C_\alpha=\varnothing$, then for open subsets of $C$ which we denote $U_\alpha=C\setminus C_\alpha$ we have $\cup_{\alpha\in I}U_\alpha=C$. Since $C$ is compact, we have finite collection $\{\alpha_1,\ldots,\alpha_n\}\subset I$ such that $C=U_{\alpha_1}\cup\ldots\cup U_{\alpha_n}$. Taking complements we get that $C_{\alpha_1}\cap\ldots C_{\alpha_n}=\varnothing$. Contradiction, so $\cap_{\alpha\in I}C_\alpha\neq\varnothing$
A family of sets with the finite intersection property is said to be centred; for convenience I will use that term.
Dan Ma’s proof is not by contradiction. He wants to prove that if every centred family of closed sets in $X$ has non-empty intersection, then $X$ is compact. To do this, he proves the contrapositive: if $X$ is not compact, then $X$ has a centred family of closed sets whose intersection is empty. This is logically equivalent to the desired implication.
The argument itself is straightforward. Suppose that $X$ is not compact; then it has an open cover $\mathscr{U}$ with no finite subcover. For each $U\in\mathscr{U}$ let $F_U=X\setminus U$, and let $\mathscr{F}=\{F_U:U\in\mathscr{U}\}$; clearly $\mathscr{F}$ is a family of closed sets. Let $\mathscr{F}_0$ be any finite subset of $\mathscr{F}$. There is a finite $\mathscr{U}_0\subseteq\mathscr{U}$ such that $\mathscr{F}_0=\{F_U:U\in\mathscr{U}_0\}$. Then
$$\bigcap\mathscr{F}_0=\bigcap_{U\in\mathscr{U}_0}F_U=\bigcap_{U\in\mathscr{U}_0}(X\setminus U)=X\setminus\bigcup\mathscr{U}_0\,.$$
$\mathscr{U}$ has no finite subcover, so $\bigcup\mathscr{U}_0\ne X$, and therefore
$$\bigcap\mathscr{F}_0=X\setminus\bigcup\mathscr{U}_0\ne\varnothing\,.$$
Thus, $\mathscr{F}$ is centred: every finite subset of $\mathscr{F}$ has non-empty intersection. But
$$\bigcap\mathscr{F}=\bigcap_{U\in\mathscr{U}}(X\setminus U)=X\setminus\bigcup\mathscr{U}=\varnothing\,,$$
since $\mathscr{U}$ is a cover of $X$, so $\mathscr{F}$ is a centred family of closed sets in $X$ whose intersection is empty.
The proof that you copied into your question uses essentially the same idea but does organize it as a proof by contradiction. I’ll try to present it a bit more clearly. We start with an arbitrary open cover $\mathscr{U}=\{U_i:i\in I\}$ of a compact space $K$, and we suppose, to get a contradiction, that it has no finite subcover. Then for each finite $J\subseteq I$ we know that $\bigcup_{j\in J}U_j\ne K$. Now for each $i\in I$ let $F_i=K\setminus U_i$; then $\mathscr{F}=\{F_i:i\in I\}$ is a family of closed sets in $K$, and for each finite $J\subseteq I$ we have
$$\bigcap_{j\in J}F_j=\bigcap_{j\in J}(K\setminus U_j)=K\setminus\bigcup_{j\in J}U_j\ne\varnothing\,,$$
so $\mathscr{F}$ is centred. We are assuming that every centred family of closed sets in $K$ has non-empty intersection, so we conclude that $\bigcap\mathscr{F}=\bigcap_{i\in I}F_i\ne\varnothing$. But then
$$\bigcup\mathscr{U}=\bigcup_{i\in I}U_i=\bigcup_{i\in I}(K\setminus F_i)=K\setminus\bigcap_{i\in I}F_i\ne K\,,$$
contradicting the fact that $\mathscr{U}$ is a cover of $K$. This contradiction shows that there must in fact be a finite $J\subseteq I$ such that $\bigcup\{U_j:j\in J\}=K$, i.e., such that $\{U_j:j\in J\}$ is a finite subcover.
Best Answer
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