Prove that an infinite chain of proper containments of compact sets is non empty

Question

I need to prove that if $K_1\supset K_2 \supset K_3 \supset K_4 \supset \ldots$ is a chain of proper containments and each $K_{i}\subseteq \mathbb{R}^{n}$ is compact, then $\bigcap_{i=1}^{\infty} K_{i} \neq \emptyset$.

I understand that $K_i \neq \emptyset, \forall i$ because of the proper subset condition and since the empty set has no proper subsets.

I also understand that the compact condition is necessary in order to prevent sets that limit, at infinity, to an empty set but for no $i$ are empty themselves (example: $K_i=(0, 1/i)$).

However, I'm not sure how to complete.

Answer 1

Pick $x_n \in K_n$. The whole sequence $(x_n)_n$ constructed this way will be in $K_1$ be the nestedness, and $K_1$ is compact. So there is an $x \in K_1$ and a convergent subsequence $(x_{n_k})_k$ that converges to $x$.

Now use the nestedness and closedness of all the $K_i$ to show that $x \in \cap_i K_i$.