The theorem, and proof presented in Rudin is:
Theorem: closed subsets of compact sets are compact
Proof: Suppose $F \subset K \subset X$, $F$ is closed (relative to $X$), and $K$ is compact. Let $\\{V_{\alpha}\\}$ be an open cover of $F$. If $F_c$ is adjoined to $\\{V_{\alpha}\\}$, we obtain an open cover $\Omega$ of $K$. Since $K$ is compact, there is a finite subcollection $\Phi$ of $\Omega$ which covers $K$, and hence $F$. If $F^c$ is a member of $\Phi$, we may remove it from $\Phi$ and still retain an open cover of $F$. We have thus shown that a finite subcollection of $\\{V_{\alpha}\\}$ covers $F$.
I read wrong, and thought the open cover $\\{ V_{\alpha}\\}$ was of $K$. Now, my question is. Why cant we just use that finite cover to proof $F$ as compact?
I know it doesnt make sense because, for example, $(0,1)$ is not compact but is a subset of $[0,1]$ which is compact. I would like to understand formally what is the problem with using the same finite subcover in $K$ and $F$.
Thank you
We know that the $V_\alpha$ cover $F$, but we cannot be sure that they cover the bigger $K$. This is why we have to adjoin $F_c$. Then we obtain an open cover of $X$ which in particular covers $K$.