For a convex set in a normed space, the closure and the weak closure coincide.
Let $S = \mathbb{sp}\{x_n\}$. Since $S$ is convex, we have $Y = \overline{S} = {\overline{S}}^w$ (the latter denoting the weak closure).
Since $x_n \to x$ weakly, we have $x \in {\overline{S}}^w$, hence $x \in \overline{S}$.
To see why $x$ must be in ${\overline{S}}^w$, suppose $x \notin {\overline{S}}^w$. Then there is a weak open set containing $x$ that does not intersect $S$. The open set must contain a weak neighborhood of the form $\{y \, |\, |\phi_i(x-y)| < \epsilon, \, \, i=1,...,n \}$. However, since $\phi_i(x_n) \to \phi(x)$, we quickly obtain a contradiction.