Weak Convergence Confusion Let $X$ a Hilbert space. Suppose there is a sequence $(x_n)$ such that for any weak neighbourhood $U$ of $0$,there is some $N$ such that for $n,m \geq N$, $x_n - x_m \in U$.
I have two questions:
(a) Is $(x_n)$ a cauchy sequence? Here is my attempt at a proof:
For any open ball around $0$, $B(\varepsilon)$,there is some weak open neighbourhood of $0$, $U$, depending on $\varepsilon$, contained in $B(\varepsilon)$. Then there is some $N$ such that for any $n,m \geq N$,
\begin{equation}
x_n - x_m \in U \subseteq B(\varepsilon).
\end{equation}
So $\|x_n - x_m\| \leq \varepsilon$, and so $(x_n)$ is a cauchy sequence (in norm topology).
(b) For such a sequence $(x_n)$, I think I should be able to show that given $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ so that for any $n,m \geq N$ and $y\in H$, $\langle x_n - x_m,y \rangle < \varepsilon$. It seems reasonable, but I have not managed to do it rigourously. 
Any help for either (or both) questions would be very appreciated. 

The condition you're given is otherwise known as weakly Cauchy.

For any open ball around $0$, $B(\varepsilon)$,there is some weak open neighbourhood of $0$, $U$, depending on $\varepsilon$, contained in $B(\varepsilon)$.

That's not true.  In fact, given a weakly open neighborhood, there is no ball that contains it.
To get a counterexample, you might try to show that any weakly convergent sequence is weakly Cauchy.  You hopefully know an example of a weakly convergent sequence that does not converge in norm.  In particular, it cannot be Cauchy in norm (by completeness).

For such a sequence $(x_n)$, I think I should be able to show that given $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ so that for any $n,m \geq N$ and $y\in H$, $\langle x_n - x_m,y \rangle < \varepsilon$. It seems reasonable, but I have not managed to do it rigourously. 

The set $\{x : \langle x ,y \rangle < \varepsilon\}$ is itself weakly open, i.e. it's a weak open neighborhood of $y$.  That is pretty much the very definition of the weak topology.  So that should help a lot.
