Proof of the finite intersection property characterisation of compactness

Proof

We pass to complements and use compactness via open covers. We read "finitely many indices" as allowing none. For no index, the intersection $K \cap F_{i_1} \cap \cdots \cap F_{i_n}$ is $K$ itself.

1. Let $X$ be a topological space, let $K \subseteq X$, and let $(F_i)_{i \in I}$ be closed subsets of $X$. Let $U_i = X \setminus F_i$, an open set.
2. Claim: for each $J \subseteq I$, $K \subseteq \bigcup_{j \in J} U_j$ holds if and only if $K \cap \bigcap_{j \in J} F_j = \emptyset$. For $J = \emptyset$, the empty union is $\emptyset$ and $K \cap \bigcap_{j \in \emptyset} F_j$ means $K$.
3. Proof of the claim: a point $x \in K$ lies outside $\bigcap_{j \in J} F_j$ if and only if $x \notin F_j$ for some $j \in J$. This means $x \in U_j$ for some $j \in J$. So each point of $K$ lies in $\bigcup_{j \in J} U_j$ if and only if no point of $K$ lies in $\bigcap_{j \in J} F_j$.
4. Assume $K$ is compact. Let $(F_i)$ satisfy $K \cap F_{i_1} \cap \cdots \cap F_{i_n} \ne \emptyset$ for all finitely many indices.
5. Suppose $K \cap \bigcap_{i \in I} F_i = \emptyset$. By step 2 with $J = I$, the open sets $U_i$ cover $K$.
6. By compactness via open covers, there are finitely many indices $i_1, \ldots, i_n$ with $K \subseteq U_{i_1} \cup \cdots \cup U_{i_n}$.
7. By step 2 with $J = \set{i_1, \ldots, i_n}$, this gives $K \cap F_{i_1} \cap \cdots \cap F_{i_n} = \emptyset$. This contradicts step 4.
8. So $K \cap \bigcap_{i \in I} F_i \ne \emptyset$, and $K$ has the property.
9. Now assume $K$ has the property. Let $(U_i)_{i \in I}$ be open sets of $X$ that cover $K$. Let $F_i = X \setminus U_i$, a closed set.
10. Suppose no finite subfamily of $(U_i)$ covers $K$. By step 2, $K \cap F_{i_1} \cap \cdots \cap F_{i_n} \ne \emptyset$ for all finitely many indices.
11. By the property, $K \cap \bigcap_{i \in I} F_i \ne \emptyset$. By step 2 with $J = I$, the sets $U_i$ do not cover $K$. This contradicts step 9.
12. So some finite subfamily of $(U_i)$ covers $K$. By compactness via open covers, $K$ is compact.