I came across an exercise that asked to characterize compactness in terms of closed sets. This is what I came up with:
Claim: $X$ is compact $\Leftrightarrow$ for every set of closed sets $\{C_\alpha\}$ with $\cap_{\alpha}C_\alpha=\emptyset$ has a finite subset $\{C_{\alpha_1},\dots,C_{\alpha_n}\}$ s.t. $C_{\alpha_1}\cap\dots\cap C_{\alpha_n}=\emptyset$.
But apparently the correct characterization is: $X$ is compact $\Leftrightarrow$ any collection of closed sets with the finite intersection property has non-empty intersection.
That seems like a much more complicated way to do it. So my question is, what's wrong with my characterization? Is mine wrong? If so, where's the mistake in the following proof? If mine is not wrong, why don't I find this characterization anywhere, while I find the other version everywhere? Thank you for any help!
Proof of Claim: $(\Rightarrow)$ Suppose $X$ is compact. Let $\{C_\alpha\}$ be a collection of closed sets s.t. $\cap_{\alpha}C_\alpha=\emptyset$. Then $\{C_\alpha^c\}$ is a collection of open sets and $\cup_\alpha C_\alpha^c=(\cap C_\alpha)^c=\emptyset^c=X$. Thus $\{C_\alpha^c\}$ is an open cover of $X$. Since $X$ is compact there is a finite subcover $C_{\alpha_1}^c, \dots, C_{\alpha_n}^c$. Since $C_{\alpha_1}^c\cup \cdots\cup C_{\alpha_n}^c=X$, $\emptyset=X^c=(C_{\alpha_1}^c\cup \cdots\cup C_{\alpha_n}^c)^c=C_{\alpha_1}\cap \cdots\cap C_{\alpha_n}$.
$(\Leftarrow)$ Let $\{U_\alpha\}$ be an open cover of $X$. Then $\{U_\alpha^c\}$ is a collection of closed sets s.t. $\cap_\alpha U_\alpha^c=(\cup U_\alpha)^c=X^c=\emptyset$. Thus $\exists$ finite subset $\{U_{\alpha_1}^c,\dots,U_{\alpha_n}^c\}$ s.t. $U_{\alpha_1}^c \cap\dots \cap U_{\alpha_n}^c=\emptyset$. But then $(U_{\alpha_1}^c \cap\dots \cap U_{\alpha_n}^c)^c=U_{\alpha_1} \cup\dots \cup U_{\alpha_n}=X$. So $\{U_\alpha\}$ has a finite subcover.