Theorem 2.36 of Baby Rudin states:
If $\{K_\alpha\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_\alpha\}$ is nonempty then $\cap K_\alpha$ is nonempty.
Rudin proves this by contradiction:
Choose $K_1 \in K_\alpha$, assume no point of $K_1$ belongs to every $K_\alpha$. Then the sets $K_\alpha^c$ form an open cover of $K_1$. Since $K_1$ is compact there are finitely many indices $\alpha_1, \dots, \alpha_n$ such that $K_1 \subset K_{\alpha_1}^c \cup \dots \cup K_{\alpha_n}^c$ which means $K_1 \cap K_{\alpha_1} \cap \dots \cap K_{\alpha_n} = \emptyset$. Hence we have an empty intersection of a finite subcollection of $K_\alpha$ which contradicts our assumption.
My question is this: is there a way to prove this directly instead of by contradiction? So to derive the fact that $\cap K_\alpha$ is nonempty directly from an assumption that every finite subcollection is nonempty.
To be clear, I understand Rudin's proof, but I am looking for alternative proof that is done directly as opposed to by contradiction. Someone marked this as a duplicate, but I do not see any existing question that addresses this specific concern.
Best Answer
Well you can start by redefining the concept "compact" by stating that a space $X$ is compact if for every family $(F_{\alpha})_{\alpha\in A}$ of closed sets that has the so-called "finite intersection property" the intersection $\bigcap_{\alpha\in A}F_{\alpha}$ is not empty.
Here family $(F_{\alpha})_{\alpha\in A}$ has by definition this "finite intersection property" if for every finite $B\subseteq A$ the intersection $\bigcap_{\alpha\in B}F_{\alpha}$ is not empty.
Using this underlying definition of "compact" (equivalent with the definition that states that open covers must have finite subcovers) it is enough to prove here that in metric spaces every compact subspace $K$ is closed (so that the family of compact sets $(K_{\alpha})$ can be recognized as a family of closed sets).
edit:
The following statements are equivalent:
(1)$\implies$(2)
Let $(F_{\alpha})_{\alpha\in A}$ be a collection of closed sets that has the finite intersection property.
Now define $U_{\alpha}=F_{\alpha}^{\complement}$ for $\alpha\in A$ and observe that for every finite $B\subseteq A$ we have $$\bigcup_{\alpha\in B}U_{\alpha}=\bigcup_{\alpha\in B}F_{\alpha}^{\complement}=(\bigcap_{\alpha\in B}F_{\alpha})^{\complement}\neq X$$
This tells us that the collection $(U_{\alpha})_{\alpha\in A}$ has no finite subcover of $X$. It is a collection of open sets, so this allows us to conclude that $(U_{\alpha})_{\alpha\in A}$ does not cover $X$.
That means that: $$\bigcap_{\alpha\in A}F_{\alpha}=\bigcap_{\alpha\in A}U_{\alpha}^{\complement}=(\bigcup_{\alpha\in A}U_{\alpha})^{\complement}\neq\varnothing$$as was to be shown.
(2)$\implies$(1)
Let $(U_{\alpha})_{\alpha\in A}$ be a collection of open sets that cover $X$.
Now define $F_{\alpha}=U_{\alpha}^{\complement}$ for $\alpha\in A$ and observe that: $$\bigcap_{\alpha\in A}F_{\alpha}=\bigcap_{\alpha\in A}U_{\alpha}^{\complement}=(\bigcup_{\alpha\in A}U_{\alpha})^{\complement}=\varnothing$$
That implies the existence of a finite $B\subseteq A$ such that $\bigcap_{\alpha\in B}F_{\alpha}=\varnothing$ (non-existence of such finite $B$ should contradict that closed family $(F_{\alpha})_{\alpha\in A}$ has the finite intersection property).
Then $$\bigcup_{\alpha\in B}U_{\alpha}=\bigcup_{\alpha\in B}F_{\alpha}^{\complement}=(\bigcap_{\alpha\in B}F_{\alpha})^{\complement}=\varnothing^{\complement}=X$$
Proved is now that cover $(U_{\alpha})_{\alpha\in A}$ has a finite subcover $(U_{\alpha})_{\alpha\in B}$ and we are ready.