Problem: Let $K$ be a compact subset of $ \mathbb R^d$ and let $A_1 \subset A_2 \subset \cdots \subset A_k$ be a nested downward sequence of closed subsets of $\mathbb R^d$. Show that if $A_k \cap K \neq \emptyset $ for each $k$, then $(\bigcap A_k) \cap K \neq \emptyset$.
attempt at solution: We will prove by contradiction. Thus suppose on the contrary that $(\bigcap A_k) \cap K = \emptyset.$ Suppose that $A_k \cap K \neq \emptyset$ for each $k$.
My idea is to use the definition of compactness to show that both expressions don't work. But I'm not exactly sure where to start with using compactness and what type of thing I should expect to be the contradiction.
Hints and Advice would be greatly appreciated.
Firstly, you need a descending sequence, thus $A_1 \supset A_1 \supset A_3\supset \cdots$ with $\text{“}\supset\text{''}$ rather than $\text{“}\subset\text{''}.$
These subsets are closed, so their respective complements $B_1 \subset B_2\subset B_3\subset \cdots$ are open.
If no point in $K$ is in all of $A_k,\, k = 1,2,3,\ldots,$ then every point in $K$ is in one of $B_k,\,k=1,2,3\ldots;$ that is de Morgan's law. Thus $\{B_k : k=1,2,3,\ldots\}$ is an open cover of $K,$ and by compactness, has a finite subcover $\{B_{k_1}, \ldots, B_{k_m}\},$ and we may take the indices to be in increasing order: $k_1<\cdots<k_m.$ Since every point in $K$ is in one of $B_{k_1},\ldots,B_{k_m},$ no point in $K$ is in $A_{k_1},\ldots, A_{k_m}.$ Thus we have $$ A_{k_1}\cap\cdots\cap A_{k_m} = \varnothing. $$ But since $A_{k_1} \supset \cdots \supset A_{k_m},$ we then have $A_{k_m} = \varnothing,$ contrary to a hypothesis.