Proposition $:$ Let $C$ be a convex and norm closed subset of a Banach space $V.$ Then $C$ is weakly closed.
Proof $:$ Let, $x_0 \notin C.$ Then by Hahn-Banach theorem there exists $f \in V^{\ast}, \alpha \in \mathbb R$ such that $$f(x_0) \lt \alpha \lt f(x)$$ for all $x \in C.$ Let $$U : = \{x \in V\ |\ f(x) \lt \alpha \}.$$ Then $U$ is $\color{blue}{\text{weakly open}}$ and $U \cap C = \varnothing.$ This shows that $C^{\complement}$ is weakly open and hence $C$ is weakly closed. $\square$
In the above proof I failed to understand why $U$ is weakly open. Instead what I know is that the set $W = \{x \in V\ |\ |f(x)| \lt \alpha \}$ is weakly open. Is there any way to say from this stage that $U$ is also weakly open?
Any help would be appreciated!
The weak topology is, by definition, the weakest topology such that all bounded linear functionals on $V$ are continuous. Therefore $f$ is continuous with respect to that topology. Observe that $$U=\{x\in V\,:\, ({\rm Re}\, f)(x)<\alpha\}=f^{-1}((-\infty,\alpha)\times \mathbb{R})$$ Therefore $U$ is open in the weak topology.