I have a problem, but first I would like to understand your statement. The statement is about a subset $C\subset l^{\infty}$ that is convex and closed in $C(l^{\infty}, |.|_{sup})$. I know that if $C$ is convex, then $C$ is closed in the weak topology. So far, so good. The problem now asks to show that C is not closed in the weak* topology.
So, how can I say that $C\subset l^{\infty}$ is closed in weak* topology if this topology is defined in its dual $l^{\infty '}$?
With this well-known, I would like a tip to show that $$C = \{x =(x_i)_{i\in \mathbb{N}} \in l^{\infty} : 0\leq x_i \leq 1, x_i \to 1 \}$$ is not closed in the topology weak*.