Property of compact subsets 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 $\alpha_1, \alpha_2, \dots, \alpha_n$ such that $K_1\subset G_{\alpha_1}\cup \dots \cup G_{\alpha_n}.$ But this means that $K_1\cap K_{\alpha_1}\cap \dots \cap K_{\alpha_n}$ is empty, in contradiction to our hypothesis.
Can you explain some moments:
1) "Assume that no point of $K_1$ belongs to every $K_{\alpha}$." What does it mean? After all, $K_1\subset \{K_{\alpha}\}$. 
2) Why $K_1\cap K_{\alpha_1}\cap \dots \cap K_{\alpha_n}$ is empty?
 A: The first point means, basically, "Assume the intersection of all $K_\alpha$ is empty". What is meant, more literally, is this: "Assume that for every $p \in K_1$, there is an $\alpha_p$ such that $p \notin K_{\alpha_p}$."
For the second point, we have that $K_1 \subseteq G_{\alpha_1}\cup \cdots \cup G_{\alpha_n}$. Taking the complement, this means that $K_1\cap (G_{\alpha_1}\cup \cdots \cup G_{\alpha_n})^c = \emptyset$. But $(G_{\alpha_1}\cup \cdots \cup G_{\alpha_n})^c = K_{\alpha_1}\cap \dots \cap K_{\alpha_n}$, so we have $K_1\cap K_{\alpha_1}\cap \dots \cap K_{\alpha_n} = \emptyset$.
A: No, $K_1$ is not contained in $\{K_\alpha \}$, it is one of its elements. 
As to ‘no point of $K_1$ belongs to every $K_\alpha$’, if it the contrary that is assumed – namely there is a point in $K_1$ that belongs to every  $K_\alpha$, the inetersection of these is non empty.
A: The way I like to think about this proof is:

A topological space is compact if every open cover has a finite subcover.

Writing this out more explicitly:

A topological space is compact if and only if whenever $(U_\alpha)$ is a collection of open sets such that $\bigcup_\alpha U_\alpha=X$, then there is a finite subcollection $\left(U_{\alpha(j)}\right)$ such that $\bigcup_{j=1}^n U_{\alpha(j)}=X$

Equivalently (contrapositive):

A topological space is compact if and only if whenever $(U_\alpha)$ is a collection of open sets such that $\bigcup_{j=1}^n U_{\alpha(j)}\ne X$ for all finite subcollections $\left(U_{\alpha(j)}\right)$, then $\bigcup_\alpha U_\alpha\ne X$

Finally, passing to complements (setting $L_\alpha=X\setminus U_\alpha$):

A topological space is compact if and only if whenever $(L_\alpha)$ is a collection of closed sets such that $\bigcup_{j=1}^n L_{\alpha(j)}\ne \emptyset$ for all finite subcollections $\left(L_{\alpha(j)}\right)$, then $\bigcup_\alpha L_\alpha\ne \emptyset$

