Okay, even assuming that $Y$ separates points the answer is still no. Take $X = l^1$ and let $Y$ be the set of elements $(a_n)$ of $l^\infty$ which satisfy $a_n \to 2a_1$. It's easy to see that $Y$ separates the points of $X$, but $e_n \to 2e_1$ weakly where $(e_n)$ is the standard basis of $l^1$. So $2e_1$ is in the weak closure of the unit ball.