Confusion in Theorem 2.36 in Rudin's PMA I shall present theorem and its proof by Walter Rudin.
Theorem. 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.
Proof: Fix a member $K_1$ of $\{K_\alpha\}$ and put $G_\alpha=K_\alpha^c$. Assume that no point of $K_1$ belongs to every $K_\alpha$. Then the sets $G_\alpha$ form an open cover of $K_1$; and since $K_1$ is compact, there are finitely many indices  $\alpha_1,…,\alpha_n$ such that $K_1\subset \cup_{i=1...n} G_{\alpha_i}$. But this means that $K_1\cap K_{\alpha_1}\cap\dots \cap K_{\alpha_n}$ is empty, in contradiction to our hypothesis.
My confusion is in the part in bold. Does mean it $K_1\cap K_{\alpha} = \emptyset $ for every $\alpha$? If the answer is "Yes", this not contradict that the intersection of every finite subcollection of $\{K_\alpha\} $ is nonempty?
 A: It means assume that no point of $K_1$ belongs to $\bigcap_{\alpha}K_{\alpha}$. ("Belongs to every $K_{\alpha}$" means "belongs to the intersection of the $K_{\alpha}$".) That means that the complements $G_{\alpha}$ of the $K_{\alpha}$ are open sets that cover the compact set $K_1$ and then $\ldots$
A: No. It means that if you fix $K_1,$ then for each $x\in K_1$ there will exist some $\alpha(x)=\alpha$ (which depends on $x$) such that $x\notin K_\alpha.$ Therefore, for each $x\in K_1$ the open set (a compact set is closed and thus its complement is open) $G_{\alpha(x)}$ contains $x$ and hence $\bigcup\limits_{\alpha(x)}G_{\alpha(x)}\supseteq K_1,$ and since $\bigcup\limits_{\alpha(x)}G_{\alpha(x)}\subseteq\bigcup\limits_{\alpha}G_\alpha $ then $\{G_\alpha\}$ forms an open cover of $K_1$ and all else I think you can understand it
A: I think I understand where your confusion lies. It is not the fact that $K_1 \cap K_\alpha = \emptyset$ $\forall \alpha$ that leads to the contraditiction. Rather, it is the fact that we can find a finite number of $\alpha_i$'s, $\alpha_1$ through $\alpha_n$, such that for these $\alpha_i$'s, $K_1 \cap K_{\alpha_i} = \emptyset$
Think of the proof in the following way:
We want to show that if every finite subcollection of $\{K_\alpha\}$ is nonempty, then $\cap^{\alpha}K_\alpha$ is nonempty.
Take any collection $\{K_\alpha\}$ where every finite subcollection of ${K_\alpha}$ is nonempty.
Assume, for the sake of contradiction, that $\cap^{\alpha}K_\alpha$ is empty. This means that there exists a $K_\alpha$, call it $K_1$, such $\textbf{that no point of $K_1$ belongs to every $K_\alpha$}$ (this is the phrase you had in bold).
All we need to do now, is to reach a contradiction. If we reach one, then that means that $\cap^{\alpha}K_\alpha$ is nonempty.
In order to reach a contradiction, Rudin finds a finite subcollection $K_1$ through $K_{{\alpha}_{n}}$ which is empty (I won't go through the explanation of how Rudin shows this because the other answers in this thread explain it clearly). This is a contradiction because it contradicts the assumption that every finite subcollection of $K_\alpha$ is nonempty.
