Cantor's intersection theorem Wikipedia proof This is the Wikipedia proof of Cantor's intersection theorem:  

$C_0\supseteq C_1\supseteq C_2...C_k\supseteq C_{k+1}$
  so that this true $\bigcap_\limits{k}^{}C_k\neq\emptyset$
  Assume, by way of contradiction, that $\bigcap_\limits{n}^{}C_k\neq\emptyset$.For each $n$, let $U_n=C_0\setminus C_n$ Since      ${\displaystyle \bigcup U_{n}=C_{0}\setminus \bigcap C_{n}} $ and ${\displaystyle \bigcap C_{n}=\emptyset }$  $\bigcap C_{n}=\emptyset$ , thus    ${\displaystyle \bigcup U_{n}=C_{0}}  {\displaystyle \bigcup U_{n}=C_{0}}$.
  Since $  C_{0}\subset S $ is compact and   ${\displaystyle (U_{n})} $ is an open cover of it, we can extract a finite cover. Let      ${\displaystyle U_{k}}$   be the largest set of this cover; then      ${\displaystyle C_{0}}\subset       {\displaystyle U_{k}} $. But then   ${\displaystyle C_{k}=C_{0}\setminus U_{k}=\emptyset }$  , a contradiction.$\blacksquare$

I want to know how $U_k$ happens to be a cover of $C_0$ how is ${\displaystyle C_{0}}\subset       {\displaystyle U_{k}} $ instead of ${\displaystyle C_{0}}=       {\displaystyle U_{k}} $ Thanks for reading!
 A: I'll give a more detailed version.
Suppose that $C_0 \supseteq C_1 \supseteq C_2 \supseteq \ldots C_{k} \ldots \supseteq C_{k+1} \ldots$, where all $C_k$ are compact non-empty (and thus closed, as we are in the reals).
Suppose for a contradiction that $\bigcap_n C_n = \emptyset$. The idea is to use that $C_0$ is compact, so we define an open cover of $C_0$ by setting $U_k = C_0 \setminus C_k$ for $k \ge 1$. Note that these are open in $C_0$ as $C_0 \setminus C_k = C_0 \cap (X \setminus C_k)$ is a relatively open subset of $C_0$ (using that all $C_k$ are closed so have open complement).
Also $U_1 \subseteq U_2 \subseteq U_3 \ldots U_k \subseteq U_{k+1} \ldots$, as the $C_k$ are decreasing.
Take $x \in C_0$. Then there is some $C_k$ such that $x \notin C_k$ (or else $x \in \bigcap_n C_n = \emptyset$), and so this $x \in U_k$ for that $k$.
This shows that the $U_n$ form an open cover of $C_0$, so finitely many $U_k$, say $U_{k_1}, U_{k_2},\ldots, U_{k_m}, k_1 < k_2 \ldots k_m$ cover $C_0$, so using the increasingness, we see hat $C_0 \subseteq U_{k_m}$. But take any $p \in C_{k_m}$ (by non-emptiness), then $p \in C_0$ and $p \notin U_{k_m}$, contradiction. So $\bigcap_n C_n \neq \emptyset$.
A: More general, suppose that $C_0\supseteq C_1\supseteq \cdots \supseteq C_k\cdots$, where all $C_k$ are compact non-empty sets. Suppose  that $\cap C_k=\emptyset$. Taking the $C_0$ complementar in both side, we have $\cup (C_0-C_k) =C_0$. Since that each set $C_i$ is closed (suppose for instance that $C_0$ is Hausdorff space), exists finite subcover $(C_0-C_{k_1}),\cdots, (C_0-C_{k_n})$ of $C_0$. Thus, $C_0=\cup_{i=1}^n(C_0-C_{k_i})$ implies that $\emptyset=\cap_{i=1}^nC_{k_i}$. This contradicts the fact of every set $C_i$ is non emptyset. 
