In the Hilbert space ($\mathcal{H}$) framework, the following Lemma holds:
Lemma: Let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $\mathcal{H}$ and let $C$ be a nonempty subset of $\mathcal{H}$. Suppose that, for every $x\in C$, $(||x_n-x||)_{n\in\mathbb{N}}$ converges and that every weak sequantial cluster point of $(x_n)_{n\in\mathbb{N}}$ belongs to $C$. Then $(x_n)_{n\in\mathbb{N}}$ converges to a point in C.
Proof: By assumption $(x_n)_{n\in\mathbb{N}}$ is bounded. A sequence like this converges in $\mathcal{H}$ if and only if it is bounded and possesses at most one weak sequential cluster point. Therefore it is enough to show that $(x_n)_{n\in\mathbb{N}}$ cannot have two distinct weak sequential cluster point in $C$. To this end, let $x$ and $y$ be weak sequential cluster points of $(x_n)_{n\in\mathbb{N}}$ in $C$, say $x_{k_n}\rightharpoonup x$ and $x_{l_n}\rightharpoonup x$. Since $x$ and $y$ belong to $C$, the sequences $(||x_n-x||)_{n\in\mathbb{N}}$ and $(||x_n-y||)_{n\in\mathbb{N}}$ converge. In turn, since: $$(\forall n \in \mathbb{N})\quad 2<x_n|x-y>=||x_n-y||^2-||x_n-x||^2+||x||^2+||x||^2-||y||^2$$ $(<x_n|x-y>)_{n\in\mathbb{N}}$ converges as wel, say $<x_n|x-y>\rightarrow \ell$. Passing to the limit along $(x_{k_n})_{n\in\mathbb{N}}$ and along $(x_{l_n})_{n\in\mathbb{N}}$ yields, respectively, $\ell = <x|x-y>=<y|x-y>$. Therefore $||x-y||^2=0$ and hence $x=y$.
If the considered space is instead a reflexive Banach space (e.g. the finite dimensional space $(\ell^n_\infty,||\cdot||_\infty)$), is it possible to achieve a similar result? Thank you in advice.