why $K_1 \cap K_{\alpha_1} \cap ...\cap K_{\alpha_n}=\emptyset$ give contradiction? I have some confusion in Baby Rudin book
My doubt is given below marked in red line and red circle

My attempt : Here we have already assume  that no point
of $K_1$ belongs to every $ K_{\alpha} $ that mean it implies already $K_1 \cap K_{\alpha_1} \cap ...\cap K_{\alpha_n}=\emptyset$
My doubts:  why  $K_1 \cap K_{\alpha_1} \cap ...\cap K_{\alpha_n}=\emptyset$ give contradiction ?
Edit:here $K_1$ is compact  and open cover are  $G_\alpha$ and so $K_1 \cap G_\alpha^c = \emptyset$ this implies  again $K_1 \cap K_\alpha=\emptyset$  ......why  its  contradicts?
 A: The key of the theorem is to get that precise contradiction, since you were assuming every finite subcollection of the compact sets $\{K_\alpha\}$ had nonempty intersection.
The way to get the contradiction is by assuming we don't have the result (reductio ad absurdum). Assuming that we don't have the result is assuming $\bigcap K_\alpha=\emptyset$.
If you fix one compact set $K_1$ then for every point $x\in K_1$ there is some $\alpha_x$ such that $x\notin K_{\alpha_x}$, since otherwise we would have $x\in K_\alpha\;\forall \alpha$, but that would mean $x\in\bigcap K_\alpha$, and we were assuming the contrary.
Then for every $x\in K_1$ there is some $\alpha_x$ such that $x\in G_{\alpha_x}:=K_{\alpha_x}^\mathsf c$. Every $G_\alpha$ is open, since compact subsets are closed in metric spaces. In consequence:
$\forall x\in K_1\;x\in G_{\alpha_x}\Rightarrow K_1\subset\bigcup_\limits{x\in K_1}G_{\alpha_x}$, so $\{G_{\alpha_x}\}_{x\in K_1}$ is an open cover of $K_1$. Since $K_1$ is compact, we must have a finite subcover $K_1\subset G_{\alpha_{x_1}}\cup\cdots\cup G_{\alpha_{x_n}}$.
Now, remember that $G_\alpha:= K_\alpha^\mathsf c$, so we actually have $K_1\subset K_{\alpha_{x_1}}^\mathsf c\cup\cdots\cup K_{\alpha_{x_n}}^\mathsf c$, which means
$K_{\alpha_{x_1}}\cap\cdots\cap K_{\alpha_{x_n}}=\big(K_{\alpha_{x_1}}^\mathsf c\cup\cdots\cup K_{\alpha_{x_n}}^\mathsf c\big)^\mathsf c\subset K_1^c$, which implies
$K_{\alpha_{x_1}}\cap\cdots\cap K_{\alpha_{x_n}}\cap K_1=\emptyset$
But $\{K_{\alpha_{x_1}},\cdots, K_{\alpha_{x_n}},K_1\}$ is a finite subcollection of the compact sets $\{K_\alpha\}$, so it should have nonempty intersection by hypothesis. We got our contradiction!
This must means that the thing we assumed is false, so the opposite is true: We have that $\bigcap K_\alpha\neq\emptyset$.
A: You have
$$
K_1\subseteq X\setminus(K_{\alpha_1}\cap K_{\alpha_2}\cap\dots\cap K_{\alpha_n})
$$
Now it's a general feature of sets that, if $A$ and $B$ are subsets of $C$, then $A\subseteq C\setminus B$ implies $A\cap B=\emptyset$, because if $a\in A$, then $a\in X\setminus B$ and therefore $a\notin B$.
The assumption on the family $(K_\alpha)$ is that the intersection of any finite set of members of the family is not empty, but $K_1,K_{\alpha_1},\dots,K_{\alpha_n}$ is a finite set of members of the family.

Maybe it is useful to see a different version of the proof. Fix a member $K$ of the family and the family of subsets of $K$ defined by $C_\alpha=K\cap K_\alpha$.
Here we're assuming that the given family is not empty, which also Rudin does (without mentioning it explicitly).
The family $(C_\alpha)$ is a family of closed subsets of $K$ (because in a metric space compact sets are closed) and $K$ is compact. If $\bigcap_{\alpha}C_\alpha=\emptyset$, then $(U_\alpha)$, where $U_\alpha=K\setminus C_\alpha$, is an open cover of $K$ and so it has a finite subcover:
$$
U_{\alpha_1}\cup U_{\alpha_2}\cup\dots U_{\alpha_n}=K
$$
that becomes
$$
C_{\alpha_1}\cap C_{\alpha_2}\cap\dots \cap C_{\alpha_n}=\emptyset
$$
But this means that
$$
(K\cap K_{\alpha_1})\cap (K\cap K_{\alpha_2})\cap\dots \cap (K\cap K_{\alpha_n})=\emptyset
$$
and therefore
$$
K\cap K_{\alpha_1}\cap K_{\alpha_2}\cap\dots \cap K_{\alpha_n}=\emptyset
$$
against the assumption.
