At here, Example $8.2$, there is this statement:
Consider any countable and dense subset $\{ x_n : n \in \mathbb{N} \}$ of the unit ball of $X$ and let $K = \{ \frac{x_n}{n} : n \in \mathbb{N} \} \cup \{ 0 \}$. Plainly, $K$ is (weakly) compact and $X = \overline{span(K)}$.
Question: Why $K$ is compact and and $X = \overline{span(K)}$?
To see that $X=\overline{span(K)}$, it's enough to show that $span(K)$ is dense in $X$ - so it's enough to show that $\{x_n: n\in\mathbb{N}\}\subseteq span(K)$. But this is easy, since $x_n={x_n\over n}+. . . +{x_n\over n}$ ($n$ times).
To show that $K$ is weakly compact: suppose I have an open cover $\mathcal{U}$ of $K$. Some open set $U_i\in\mathcal{U}$ contains $0$, since $0\in K$. What can you tell me about $K\setminus U_i$?