Let $E$ be a normed vector space.
In an exercise for a homework in my Functional Analysis class, my professor defined weak convergence for sequences as $(x_n)$ is weakly convergent to $x \in E$ iff for each $f \in E^*$ - the space of continuous functionals - $\lim_{n \to \infty}f(x_n) = x$. Then he proceeds defining a set $X \subseteq E$ to be weakly closed iff for each weakly convergent sequence $(x_n) \subseteq X$, its limit is an element of $X$.
Now here's where I got confused: this homework was previous to the lecture in which he properly introduced the weak topology of $E$. Bearing in mind that for each infinite dimensional normed vector space, its weak topology is never metrizable, it can't be, particularly, first countable. So the sequence criterion for characterizing closed sets is not guaranteed (and false, I suspect). So the definition of 'weakly closed set' given in the homework isn't immediately equivalent to the definition of closed set in the weak topology. Are they actually the same?
The exercise also asks us to show a subset of a Banach space that is closed in the norm topology, but isn't weakly closed (in the sense of weakly convergent sequences above). I thought about the unit sphere, as its closure in the weak topology is the closed unit ball - but I still haven't related this result to his definition of weakly closed.
