Please check my proof since I don't have a solutions manual. Thank you very much.
Let $(K_n:n\in\Bbb{N})$ be a sequence of nonempty compact sets in $\Bbb{R}$ such that $K_1\supseteq K_2\supseteq\cdots\supseteq K_n\supseteq\cdots$. Prove that there exists at least one point $x\in\Bbb{R}$ such that $x\in K_n$ for all $n\in\Bbb{N}$; that is $\bigcap_{n=1}^{\infty}K_n\neq\emptyset$.
Proof: If for all $n\in\Bbb{N}$, $K_n$ is compact, then by the Heine-Borel Theorem, $K_n$ is closed and bounded. Let $(\sup K_n)_{n=1}^{\infty}$ be a sequence of supremums of the decreasing sequence of compact sets $(K_n)$. Note that $(\sup K_n)_{n=1}^{\infty}$ is decreasing and bounded so, by the monotone convergence theorem, $(\sup K_n)_{n=1}^{\infty}$ is convergent to some $x'\in\Bbb{R}$. Consider the set \begin{align} S_j=\{\mu\in\Bbb{R}\quad\vert\quad \exists n\geq j\in\Bbb{N},\quad \sup K_n=\mu\} \end{align} , then $S_j\subseteq\bigcup_{n=j}^{\infty}K_n=K_j$. Notice that $x'$ is a point of closure of $S_j$ and thus, also of $K_j$. Since $K_j$ is closed due to compactness, then $x'\in K_j$. However, $j$ is arbitrary and therefore, for all $n\in\Bbb{N}$, $x'\in K_n$; that is $x'\in\bigcap_{n=1}^{\infty}K_n\neq\emptyset$.