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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 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.