Exercise 3.3.3

Let $K$ be closed and bounded and let $(x_n)$ be a sequence contained in $K$. BW tells us a convergent subsequence $(x_{n_k})\to x$ exists since $K$ is bounded, and since $K$ is closed $x\in K$. Thus every sequence in $K$ contains a subsequence convering to a limit in $K$, which is the definition of $K$ being compact.