Let $X$ be a Banach Space. We say that
A mapping $T: C\subset X \to C$ is nonexpansive if $\|Tx - Ty\| \leq \|x-y\|$, for all $x,y \in C$.
$X$ has the Fixed Point Property (FPP) if every nonexpansive endomorphism of a nonempty, bounded, closed and convex subset $A \subset X$ has a fixed point.
$X$ has the Weak Fixed Point Property (w-FPP) if every nonexpansive endomorphism of a nonempty, weakly compact, convex subset $A \subset X$ has a fixed point.
From what I have read, some authors claim that, in a Reflexive Space, both the w-FPP and the FPP coincide but I don't understand why.
I can see how the Eberlein–Šmulian theorem tell us that the FPP $\implies$ w-FPP and that I need to prove that in a Reflexive Space every nonempty, weakly compact and convex set is closed and bounded.
I was thinking that perhaps using the fact that a Banach Space $X$ is reflexive iff the closed unit ball is weakly compact would be helpful, however I do not know how to proceed.
Use that a convex set is norm closed iff it is weakly closed.
Using additionally that weakly compact sets are bounded, you immediately see that a weakly compact, convex subset $A\subset X$ is norm closed, bounded, and convex.
On the other hand, if a subset $A\subset X$ is convex, norm closed, and bounded, your own characterization of reflexiveness implies that $A$ is a weakly closed subset of a weakly compact set. Hence $A$ is weakly compact and convex.
This should prove both directions.