Confusion in Theorem 2.36 Baby Rudin

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 $\ldots$

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