Weakly Compact generated is if there is weakly compact subset $K$ of Banach space $\mathbb{X}$ that closed liner span is $\mathbb{X}$.
Corollary 3. A Banach space $X$ is weakly compact generated if and only if there is a reflexive Banach space $R$ and a one-to-one operator $T:R\longrightarrow X$ with $T(R)$ dense in $X$.
Proof. The sufficiency is obvious. For the necessity, let $K$ be a weakly compact subset of $X$ whose linear span is norm dense in $X$ and in Lemma $1$ let $W$ be the convex hull of $K\cup (-K)$, so that $Y$ is reflexive and $j$ gives the desired one-to-one operator. $\square$
This is from Creating a reflexive Banach space from a weakly compact set. (Philips answer)
I have trouble with sufficiency. Why this is obvious..? Is $R$ weakly compact? I think that $T$ is weakly compact operator, so $cl(T(ball R))$ is weakly compact, but it cannot deduce $R$ is weakly compact.