Equivalence of Definitions of Compact Topological Space

$(2) \implies (4)$
Let $T = \left({S, \tau}\right)$ satisfy the Finite Intersection Axiom.

Let $\mathcal F$ be a filter on $X$.

that $\mathcal F$ has no limit point.

Thus:
 * $\bigcap \left\{{\overline F : F \in \mathcal F}\right\} = \varnothing$

By hypothesis there are therefore sets $F_1, \ldots, F_n \in \mathcal F$ such that:
 * $\overline F_1 \cap \ldots \cap \overline F_n = \varnothing$

Because for any set $M$ we have $M \subseteq \overline M$:
 * $\overline F_1, \ldots, \overline F_n \in \mathcal F$

But by definition of a filter, $\mathcal F$ must not contain the empty set.

Thus $\mathcal F$ has a limit point.

$(4) \implies (2)$
Let $\mathcal A \subset \mathcal P \left({S}\right)$ be a set of closed subsets of $S$.

Let:
 * $\bigcap \tilde{\mathcal A} \ne \varnothing$

for all finite subsets $\tilde{\mathcal A}$ of $\mathcal A$.

We show that this implies $\bigcap \mathcal A \ne \varnothing$.

From our assumption, $\mathcal B := \left\{{\bigcap \tilde{\mathcal A} : \tilde{\mathcal A} \subseteq \mathcal A \text{ finite}}\right\}$ is a filter basis.

Let $\mathcal F$ be the corresponding generated filter.

By hypothesis $\mathcal F$ has a limit point.

Thus:
 * $\varnothing \ne \bigcup \left\{{\overline F : F \in \mathcal F}\right\} \subseteq \bigcap \mathcal B \subseteq \bigcap \mathcal A$.

Thus $\bigcap \mathcal A \ne \varnothing$.

Hence $T = \left({S, \tau}\right)$ satisfies the Finite Intersection Axiom.