In Hilbert spaces, it is well-known that a sequence $\{x_n\}$ is weakly convergent if and only if it is bounded and possesses at most one weak sequential cluster point.
My question is whether or not the above fact holds in uniformly convex spaces. Thank you for all the solutions.
This property holds in every reflexive Banach(1) space. A Banach space is reflexive if and only if its closed unit ball is weakly compact. Of course that holds for balls with a different radius too, a Banach space is reflexive if and only if every bounded subset is relatively weakly compact.
A bounded sequence $(x_n)_{n \in \mathbb{N}}$ is contained in a closed ball $K := \overline{B_r(0)}$ for some $r > 0$. If the space is reflexive, $K$ is weakly compact, and therefore $(x_n)$ has a weak sequential cluster point $z\in K$. If the sequence isn't weakly convergent to $z$, there is an open weak neighbourhood $U$ such that $A = \{n \in \mathbb{N} : x_n \notin U\}$ is infinite. Then $(x_n)_{n \in A}$ is a sequence in the weakly compact set $K \setminus U$, and therefore has a weak sequential cluster point $w \in K\setminus U$. Of course $w$ is also a weak sequential cluster point of the full sequence $(x_n)_{n \in \mathbb{N}}$.
Thus a bounded sequence in a reflexive Banach space that isn't weakly convergent has at least two weak sequential cluster points.
The contrapositive is that a bounded sequence in a reflexive Banach space that has at most one weak sequential cluster point must converge weakly (and hence have exactly one weak sequential cluster point).
Uniformly convex Banach spaces are reflexive.
(1) Completeness is of course necessary, a non-convergent Cauchy sequence in a normed space is bounded and has no weak sequential cluster point, hence isn't weakly convergent.