Show that the closures in the topology weak and the norm are coincide:

If $x$ is not in $Y_F$, there is a member $f^*\in X^*$ such that $f^*(x)=1$ and $f(Y_F)=\{0\}$. The weak nhood of $x$ determined by $f^*$ and $\epsilon=1/2$ contains $x$ and is disjoint from $Y_F$. It follows that $x$ is not in $Y_f$. So the weak closure of $Y$ is contained in the norm closure of $Y$.

The reverse inclusion is obvious.

Each convex subset E of locally convex space X has equal closures in original and weak topologies. See theorem 3.12 in Rudin's Functional analysis.