I really don't like the way the proof is presented in Rudin, and I'd go as follows: suppose $\;\bigcap_\alpha K_\alpha=\emptyset\;$ , but then fixing $\;\alpha_0\;$ we get (de Morgan):
$$K_{\alpha_0}=K_{\alpha_0}\setminus\emptyset=K_{\alpha_0}\setminus\bigcap_\alpha K_\alpha=\bigcup_\alpha\left(K_{\alpha_0}\setminus K_\alpha\right)$$
But $\;K_{\alpha_0}\setminus K_\alpha\;$ is open for every $\;\alpha\;$, so by compactness of $\;K_{\alpha_0}\;$ we have that there exists a finite set $\;\{\alpha_1,...,\alpha_n\}\;$ s.t.
$$K_{\alpha_0}=\bigcup_{i=1}^n\left(K_{\alpha_0}\setminus K_{\alpha_i}\right)=K_{\alpha_0}\setminus\bigcap_{i=1}^nK_{\alpha_i}\implies \bigcap_{i=1}^nK_{\alpha_i}=\emptyset$$
and there's our contradiction.
Of course, nothing new is presented above. It is just the way it is presented that changes and that imo makes it easier to grasp.
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) Every collection of open sets that cover $X$ has a finite subcover.
- (2) Every collection of closed sets that has the finite intersection propery has a non-empty intersection.
(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.
Best Answer
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 $\ldots$