 Solution

Every $K_n$ is contained in the bounded set $K_1$, so the intersection $K={\bigcap }_{n\ge 1}{K}_n$ 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 $n_j\ge m$ 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 $K\in \mathcal{K}$.