Okay, even assuming that $Y$ separates points the answer is still no. Take $X = l^1$ and let $Y$ be the set of sequences $(a_n)_n$ in $l^\infty$ which satisfy $\lim_{n\to\infty}a_n = 2a_1$. It's easy to see that $Y$ separates the points of $X$, but $e_n \to 2e_1$ weakly where $\{e_n\in l^1\}_n$ is the standard basis. So $2e_1$ is in the weak closure of the unit ball.
Nik Weaver
- 42.8k
- 3
- 112
- 213