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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.

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 $\lim a_n = 2a_1$. It's easy to see that $Y$ separates the points of $X$, but $e_n \to 2e_1$ weakly where $(e_n)$ is the standard basis of $l^1$. So $2e_1$ is in the weak closure of the unit ball.

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.

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.

Okay, even assuming that $Y$ separates points the answer is still no. Take $X = l^1$ and let $Y$ be the algebraic span of the elements $(1,2,2,2,\ldots)$, $(1, 0, 2,2,2,\ldots)$, $(1,0,0,2,2,2,\ldots)$, etc., in $l^\infty$. Any element of $l^1$ that sums to zero against each of these sequences also sums to zero against their successive differences, which easily implies that it must be zero. So $Y$ separates the points of $X$. But any finitely many of these sequences fail to separate $(2,0,0,0,\ldots) \in l^1$ from a sequence which is $1$ in a single entry sufficiently far out, and zero in all other entries. So $(2,0,0,0,\ldots)$ is in the weak closure of the unit ball.

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.