Proving $B=\{x\in E:\|x\|<1\}$ is not open under the weak topology 
Prove that $B=\{x\in E:\|x\|<1\}$ is not an open set in $E$ under the weak topology $\sigma(E,E^*)$.

A neighbourhood around $x_0$ in the weak topology $\sigma(E,E*)$ is given by $\mathscr{V}=V(x_0;T_1,T_2,...,T_n;\epsilon)\forall i \in \{1,2,3,\ldots,n\}$.
By Hahn-Banach theorem $\|.\|\to\mathbb{K}$ is a functional.
So regarding this functional I can take the sequence ${\{x_n\}_{n\in\mathbb{N}}}\in B $ so that it converges to $x_0\in E$. Then $\forall\epsilon>0,\exists N\in\mathbb{N}:\|x_n-x_0\|<\epsilon$
I need to prove $x_0\in B$
$\epsilon>\|x_n-x_0\|\geqslant\|x_n\|-\|x_0\|\implies \|x_n\|-\epsilon>\|x_0\|\implies 1-\epsilon>\|x_0\|$ so $x_0\in B$ and $B$ is closed regarding the functional $\|.\|\to\mathbb{K}$
I could also have applied Heine theorem since the norm is continuous.
However this is not enough once I need to prove that $B$ is not open regarding all the functionals.
Questions:
How do I solve this problem? How do I go on proving for every functional?
Thanks in advance!

The norm function is not a functional: it's not linear!
To see $B$ is not open show it has a non-interior point, I pick $0$:
Suppose $0 \in B$ has a basic open neighbourhood $$ 0 \in V(0; T_1, T_2, \ldots, T_n; \varepsilon) \subseteq B\tag{1}$$ 
Assuming, as we must, that $E$ is infinite-dimensional, we find a non-zero vector  $y_0$ in the common kernel of the $T_i$, as in your other question, then $z_0:=\frac{1}{\|y_0\|}\cdot y_0$ has norm $1$ so is not in $B$, while by construction $z_0$ being a multiple of $y_0$, has all $T_i(z_0) = 0$ and so $z_0 \in  V(0; T_1, T_2, \ldots, T_n; \varepsilon)$ but $z_0 \notin B$ contradicting the inclusion $(1)$. 
This means that $0$ is not an interior point of $B$ in the weak topology, so $B$ is not open in that topology.
