I made a mistake while proving the theorem in the title, however, I don't see where.
Here's the theorem:
Let $C$ be closed and limited subsets of $\Bbb{R}$. Then $\bigcap_{i \in K} (C_i) = \emptyset \Rightarrow \exists F \subseteq K \; \text {finite} | \bigcap_{i \in F} (C_i) = \emptyset$
Proof (proceeds by finding equivalent problems).
$$\Bbb{R} - \bigcap_{i \in K} (C_i) = \Bbb{R} \Rightarrow \exists F \subseteq K \; \text {finite} | \Bbb{R} - \bigcap_{i \in F} (C_i) = \Bbb{R}$$
$$\bigcup_{i \in K} (\Bbb{R} - C_i) = \Bbb{R} \Rightarrow \exists F \subseteq K \; \text {finite} | \bigcup_{i \in F} (\Bbb{R} - C_i) = \Bbb{R}$$
Let $O_i = \Bbb{R} - C_i$. Note that $\forall i \in K. \, O_i$ is open and unlimited subset of $\Bbb{R}$
$$\bigcup_{i \in K} (O_i) = \Bbb{R} \Rightarrow \exists F \subseteq K \; \text {finite} | \bigcup_{i \in F} (O_i) = \Bbb{R}$$
And the last is false since if we take $\forall i \in \Bbb{N}. \, O_i = (-\infty , i), B = \bigcup_{i \in \Bbb{N}} (O_i) = \Bbb{R}$ but there are no finite subsets of N which generates R in that way.
I cannot.find the flaw in my proof. Can you please help me with that?
For example you cannot take $ O_i = (-\infty,i) $ as an example because theese $ O_i $ are not the complement of a compact subset of $ \mathbb{R} $.
A proof would be as follows: Fix a $ C_1 $. Since $ \bigcap_{i \in K} C_i = \emptyset $ we have $ \forall x \in C_1 : \exists 1 \neq i \in K : x \notin C_i $. Hence $ \{ C_i^C \}_{i \in K} $ is an open covering of $ C_1 $ and we can choose $ i_1 , ... , i_n \in K$ with $ C_1 \subset \bigcup_{j=1}^{n} {C_{i_j}}^ C.$ Which implies $ C_1 \bigcap_{j=1}^{n} C_{i_j} = \emptyset $.