# If every subsequence of $(x_n)$ has a subsequence converging weakly to $x$ then $x_n$ converges weakly to $x$.

Let $H$ be a Hilbert space(or a reflexive Banach space) and $(x_n)$ a sequence in $H$. Is the following proposition true?

If every subsequence of $(x_n)$ has a subsequence converging weakly to $x$ then $x_n$ converges weakly to $x$.

I think this is true for bounded sequences since bounded sets are weakly sequencially compact. But I couldn't prove it.

Let $X$ be a topological space, $\left\{x_n\right\}\subset X$ a sequence, and $x\in X$. Then the following are equivalent:
1. $x_n$ converges to $x$;
2. any subsequence of $\left\{x_n\right\}$ admits a subsequence which converges to $x$.
Proof: $(1)\Rightarrow (2)$ is obvious. Assume $(2)$ and suppose by contradiction that $x_n\not \to x$. Then there is a neighbourhood $\mathcal{U}$ of $x$ and a subsequence $\left\{x_{n_k}\right\}\subset \left\{x_n\right\}$ such that $\left\{x_{n_k}\right\}$ stays away from $\mathcal{U}$. But by assumption $\left\{x_{n_k}\right\}$ should have a subsequence converging to $x$, which is impossible.