Motivation: trying to prove that if $K \subseteq \mathbb{R}$ is compact (and thus, by the Heine-Borel theorem, closed and bounded), then this implies that any open cover for $K$ has a finite subcover.
Exercise: In order to prove the above, I first need to show the following:
Let $\{ O_\lambda \mid \lambda \in \Lambda \}$ be an open cover for $K$ and, for contradiction, let us assume that no finite subcover exists for $K$. Let $I_0$ be a closed interval containing $K$, and bisect $I_0$ into two closed intervals $A_1$ and $B_1$. Why must either $A_1 \cap K$ or $B_1 \cap K$ (or both) have no finite subcover consisting of sets from $\{O_\lambda \mid \lambda \in \Lambda \}$?
I'm not sure how to prove the above. Since by assumption $K$ has no finite subcover, this means that: \begin{equation} K \subsetneq O_{\lambda_1} \cup O_{\lambda_2} \cup \cdots \cup O_{\lambda_n} \end{equation} Now, if we cut $I_0 \supseteq K$ into two intervals, then surely there will be a point $x \in K \subseteq I_0$, with $x \notin O_{\lambda_1} \cup O_{\lambda_2} \cup \cdots \cup O_{\lambda_n}$, such that: \begin{equation} x \in A_1 \cap K \subsetneq O_{\lambda_1} \cup O_{\lambda_2} \cup \cdots \cup O_{\lambda_n} \end{equation} or: \begin{equation} x \in B_1 \cap K \subsetneq O_{\lambda_1} \cup O_{\lambda_2} \cup \cdots \cup O_{\lambda_n} \end{equation} Does this count as a valid proof? I feel like I'm missing something, but am not sure what it is.
Your proof idea works, but the phrasing is imprecise and crucial details are left out. What you're trying to do is to show that for any finite subcollection $O_{\lambda_1},\ldots O_{\lambda_n}$,the following holds: $A\cap K\subsetneq O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$ or $B\cap K\subsetneq O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$. This amounts to showing that there is an $x\in A\cap K$ or an $x\in B\cap K$ such that $x\not \in O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$.
Now, since by assumption $K\subsetneq O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$, there is an $x\in K$ such that $x\not \in O_{\lambda_1}\cup\ldots \cup O_{\lambda_n}$. Furthermore, since $K\subset A\cup B$, we must have $x\in A$ or $x\in B$ (or both), so $x\in A\cap K$ or $x\in B\cap K$.