Let $X$ be a normed linear space and $F\subset X^*$. Let $\overline{F}$ denotes the closure of $F$ in $X^*$ in the norm topology. Give $X$ the weak topology induced by the family $F$ denoted by $\sigma(X,F)$. I need to prove that $\phi \in \overline{F}$ then $\phi$ restricted to the unit ball of $X$ is $\sigma(X,F)$ continuous. I tried to proceed like this. Since $\phi \in \overline{F}$, there exists $\phi_n$ such that $||\phi-\phi_n|| \to 0$ s $n \to \infty$. To prove $\phi$ restricted to the unit ball of $X$ is $\sigma(X,F)$ continuous, it is enough to prove any net $x_i$ in the unit ball converging to $x$, $\phi(x_i)$ converges to $\phi(x)$.
Observe that
$|\phi(x_i)-\phi(x)|\leq |\phi(x_i)-\phi_n(x_i)|+|\phi_n(x_i)-\phi_n(x)|+|\phi_n(x)-\phi(x)|$
first inequality can be made small letting $n \to \infty$ since $\phi_n \to \phi$ in norm. Similarly third inequality also. Second follows from the fact that $\phi_n \in F$ and $\sigma(X,F)$ is the smallest topology for which each element in $F$ is continuous.
But still I am not getting where I am using it. But I am not using $x_i$'s are in unit ball. Where I am getting wrong?