Let $X$ be an infinite dimensional, non-reflexive Banach space. Then I am looking for an example of a set $A \subseteq X$ which is weakly compact but not strong compact i.e. not norm compact.
Since $A$ is weakly closed it is also norm closed and since $A$ is weakly bounded, it is also bounded. Hence $A$ is necessarily bounded but not totally bounded, as otherwise $A$ were strongly compact.
Furthermore, the unit ball $B_X$ (and of course any translation and rescalings of it) is not an example: If $B_X$ is weakly compact, then the space is reflexive. This is theorem 3.31.
In $\ell^1$, no such sets exist, as seen here; basically, strong and weak convergence coincide, which is non-trivial. After that it is a consequence of Eberlein–Šmulian.
Any partial answers of references are greatly appreciated too. Thank you in advance.
David Mitras answer checks out. Thank you to him.
Consider the set $A:= \{e_n\}_{n \in \mathbb{N}} \cup \{0\} \subseteq (c_0, \Vert \cdot \Vert_{\infty})$. Then we have
$A$ is weakly compact. By Eberlein–Šmulian it is enough to show sequential compactness, so let $(x_n)_{n \in \mathbb{N}}$ be a sequence in $A$. Then it will eventually be a subsequence of $(e_n)_{n \in \mathbb{N}}$ and hence
$$ \forall y \in (c_0)^* \simeq \ell^{1}: \lim_{n \rightarrow \infty} y(x_n) = \lim_{n \rightarrow \infty} y(e_n) = \lim_{n \rightarrow \infty} \sum_{i = 1}^{\infty} y_i \delta^i_n = \lim_{n \rightarrow \infty} y_n = 0$$
since $y_n \rightarrow 0$ as $n \rightarrow \infty$.
However, the set is not norm compact since for the sequence $(e_n)_{n \in \mathbb{N}}$ we have $\forall i \neq j: B_{1/2}(e_i) \cap B_{1/2}(e_j) = \emptyset$ and has thus no convergent subsequence.