Weakly closed implies sequentially closed 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!? :(
Instead of saying "$A$ is sequentially closed", the conclusion of the problem really should say that $A$ is "weakly sequentially closed" or "sequentially weakly closed." 

The problem is asking to show that closed in the weak topology implies sequentially closed in the weak topology.  

A solution can be given that does not use the specifics of the weak topology at all. For any topological space, closed implies sequentially closed.  

Let $X$ be any topological space and let be $A$ any closed subset of $X$. We need to show that $A$ is sequentially closed. Let $x \in X$ and $(x_n) \subseteq A$ be such that $x_n \to x$. We need to show $x \in A$. Let $U$ be an arbitrary  open neighbourhood of $x$. Since $x_n \to x$, we have $x_n \in U$ for all but finitely many $n$, and so $U$ contains at least one point of $A$. Since $U$ was an arbitrary open neighbourhood of $x$, this proves $x$ is in the closure of $A$. Since $A$ is closed, $x$ is in $A$.
Can't you just say that since $A$ is weakly closed then $A$ is strongly closed, then $A$ is sequentially closed? Cause I think every weakly closed set is strongly closed and in metric spaces (we are in a normed space which means in a metric space) (strongly) being closed set is equivalent to being sequentially closed.