Another problem involving the weak topology:

Let $X$ be a normed space and $A \subset X$ weakly closed. Then $A$ is sequentially closed, that is: If $(x\_n) \subset A$ and $x\_n \xrightarrow{w}x$, then $x \in A$.

I know this characterisation is also used as definition of weakly closed. So I guess it should be easy to prove. Yet I have trouble doing so. ;(

I tried proving it directly and towards a contradiction without success. I'm afraid my problem is a lack of understanding of weak closedness. I know how weakly open sets are generated, but this doesn't give me a concrete representation them, or of a weakly closed subset.

However, I know that $A$ is also closed with respect to the norm of $X$. Weak convergence of $(x\_n)$ to $x$ means $f(x\_n) \rightarrow f(x)$ for every $f \in X^\*$. But it doesn't give me any statement related to the norm convergence of $(x\_n)$, at least not that I know of. So the norm-closedness of $A$ doesn't really help. I could also find out that norm closedness is not sufficient for closedness with respect to weak convergence - I think convexivity has to be added to make the implication valid, is that correct?

So of course I have to fail if I weak convergence with norm convergence alone.

I've tried to work with balls around $f(x)$, too (for a contradiction). But again, the continuousness of $f$ only gives me control over $|f(x\_n)-f(x)|$ if I have some bound for $||x\_n-x||$. And I want it the other way around. I have a feeling that this way is wrong because it would need some implication between weak and norm convergence that I know isn't there...

It makes me mad that the proof should be rather simple, yet I'm not able to do it. Some hint please!? :(