On the uniqueness of certain weak cluster point in Hilbert space The problem goes as follows:
Let $\mathcal{H}$ be real Hilbert space, $C\subset \mathcal{H}$ be a subset. Let $\{x_n\}\subset\mathcal{H}$ satisfies the following property:
$$||x_{n+1}-x||\leq||x_n-x||$$
holds for all $x \in C$. Show that $\{x_n\}$ has at most one weak cluster point in $C$.
I tried to use Mazur's theorem to find some convex combination of $\{x_n\}$ making the weak convergence into strong one, but then the contraction property does not hold anyway. Though it is an easy observation that $\{||x_n-x||\}$ is convergent sequence for any $\forall x \in C$, I had no idea of whether this may be of help or not.
 A: Suppose $a \in C$ is a weak cluster point of $(x_k)$. Translating by $a$, if necessary, we can assume $a = 0$. Let $b \in C,\, b \neq 0$, and $e = \frac{1}{\lVert b\rVert}\cdot b$. For $x \in\mathcal{H}$, write $x = x' + x''$ with $x'$ a multiple of $b$ and $\langle x'' \mspace{-3mu}\mid\mspace{-3mu}b\rangle = 0$, so $x' = \langle x \mspace{-3mu}\mid\mspace{-3mu}e\rangle e$.
To see that $b$ is not a weak cluster point of $(x_k)$, consider first ($\lVert x_k\rVert$ is a non-increasing sequence by the premise)
$$r = \lim_{k\to\infty} \lVert x_k\rVert = \inf \lVert x_k\rVert.$$
If $r < \lVert b\rVert$, the weak neighbourhood $\left\lbrace x : \langle x \mspace{-3mu}\mid\mspace{-3mu}e\rangle > \frac{r + \lVert b\rVert}{2}\right\rbrace$ of $b$ contains only finitely many $x_k$ (no $x_k$ with $\lVert x_k\rVert < \frac{r + \lVert b\rVert}{2}$ is contained in that weak neighbourhood).
So suppose $r \geqslant \lVert b\rVert$. For $0 < \varepsilon < \lVert b\rVert/3$, consider the weak neighbourhoods $A_\varepsilon = \{ x : \lvert\langle x \mspace{-3mu}\mid\mspace{-3mu}e\rangle\rvert < \varepsilon \}$ of $0$ and $B_\varepsilon = \{ x : \bigl\lvert\langle x \mspace{-3mu}\mid\mspace{-3mu}e\rangle - \lVert b\rVert\bigr\rvert < \varepsilon\}$ of $b$.
Since by assumption $0$ is a weak cluster point of the $x_k$, there is a $k(\varepsilon)$ with $\lVert x_{k(\varepsilon)}\rVert^2 < r^2 + \varepsilon^2$ and $x_{k(\varepsilon)} \in A_\varepsilon$.
If there is an $m > k(\varepsilon)$ with $x_m \in B_\varepsilon$, let $n > m$ with $x_n \in A_\varepsilon$. Then we have, since also the $\lVert x_k - b\rVert$ are non-increasing by the premise,
$$\begin{align}
r^2 + \varepsilon^2 &>\lVert x_m\rVert^2 = \lVert x_m'\rVert^2 + \lVert x_m''\rVert^2\\
&\geqslant (\lVert b\rVert -\varepsilon)^2 + \lVert x_m''\rVert^2\\
\Rightarrow \lVert x_m''\rVert^2 &\leqslant r^2 + \varepsilon^2 - (\lVert b\rVert-\varepsilon)^2 = r^2 - \lVert b\rVert^2 + 2\lVert b\rVert\varepsilon\\
\Rightarrow \lVert x_m-b\rVert^2 &\leqslant \lVert x_m''\rVert^2 + \varepsilon^2
\leqslant r^2 - \lVert b\rVert^2 + 2\lVert b\rVert\varepsilon + \varepsilon^2
\end{align}$$
and
$$\begin{align}
r^2 - \lVert b\rVert^2 + 2\lVert b\rVert\varepsilon + \varepsilon^2
&\geqslant \lVert x_n -b\rVert^2 = \lVert x_n'-b\rVert^2 + \lVert x_n''\rVert^2\\
&\geqslant \bigl(\lVert b\rVert -\varepsilon\bigr)^2 + \lVert x_n''\rVert^2
\end{align}$$
whence
$$\lVert x_n''\rVert^2 \leqslant r^2 - 2\lVert b\rVert^2 + 4\lVert b\rVert\varepsilon$$
and
$$\lVert x_n\rVert^2 \leqslant \lVert x_n''\rVert^2 + \varepsilon^2 \leqslant r^2 - \bigl(2\lVert b\rVert^2 - 4\lVert b\rVert\varepsilon - \varepsilon^2\bigr) < r^2,$$
which contradicts $r = \inf \lVert x_k\rVert$.
Thus $x_k \notin B_\varepsilon$ for $k \geqslant k(\varepsilon)$, and hence $b$ is not a weak cluster point of $(x_k)$. 
A: Considering you have two distinct cluster points in $C$, $a$ and $b$. Without loss of generality $\Arrowvert a - b \Arrowvert = 1$
Let $n$ such as $\Arrowvert a-x_{n} \Arrowvert <1/3$. 
Let $m > n$ such as $\Arrowvert x_{m}-b\Arrowvert < 1/3$
\begin{eqnarray}
\Arrowvert a-b \Arrowvert & \leq & \Arrowvert a-x_{m} \Arrowvert +\Arrowvert b-x_{m}\Arrowvert \; \mbox{triangular inequality} \\
\Arrowvert a-x_{m} \Arrowvert & \geq & 2/3 \geq \Arrowvert a- x_n \Arrowvert
\end{eqnarray}
Contradicting your hypothesis 

for all $x\in C$, $\Arrowvert x_{n+1} - x \Arrowvert \leq \Arrowvert x_n - x \Arrowvert$

