= Solution
Every $K_n$ is contained in the bounded set $K_1$, so the <intersection> $K=\bigcap_{n\geq1}K_n$ is bounded. It is closed because an arbitrary intersection of <closed sets> is closed.
It remains to prove that $K$ is nonempty. Choose $x_n\in K_n$. The set $K_1$ is a <compact set>, so some <subsequence> $x_{n_j}$ converges to a point $x\in K_1$. For each fixed $m$, all sufficiently late terms satisfy $n_j\geq m$ and hence $x_{n_j}\in K_{n_j}\subseteq K_m$. Since $K_m$ is closed, $x\in K_m$. This holds for every $m$, so $x\in K$. Thus $K\in\mathcal K$.
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