As usual I struggle with the weak topologies...
There's a probably silly detail I'm missing in "Rudin's Functional analysis" section 3.11. "The weak topology of a topological vector space".
At some point it is stated:
... Since every weak neighborhood of $0$ contains a neighborhood of the form $$ V = \left\{ x : |\Lambda_i x| < r_i \;\text{for}\; 1 \leq i \leq n \right\} $$ Where $\Lambda_i \in X^*,r_i > 0$ it is easy to see that $x_n\to 0 $ weakly if and only if $\Lambda x_n \to 0$ for every $\Lambda \in X^*$.
The first bit is how to prove that every weak neighborhood contains sets of the form $V$. As far as I understood every weak neighborhood is defined as union of of finite intersections of sets $\Lambda^{-1}(U)$, with $\Lambda \in X^*$ and $V$ open in the field that defines the topological vector space $U$.
I suppose since I can read $V$ as finite intersections of counter images of open sets in the field then by definition of weak topology every weak neighborhood contains sets of the form $V$.
I think the argument above is correct, but I'd like to see it rigorously.
The other questions is how do I prove $x_n \to 0$ iff $\Lambda x_n \to 0$ for every $\Lambda \in X^*$?
If $V$ is a weak neighborhood of $0$ then $\cap_k \{x:\Lambda_k(x)^{-1}(U_k)\} \subset V$ for some finite set of continuous linear maps $\Lambda_k$ and some open sets $U_k$ in the scalar field. Since $0$ is in this set we have $0=\Lambda_k(0)\in U_k$. Hence there exists $r_k >0$ such that $|c| <r_k$ implies $c \in U_k$. Now $\cap_k \{x:|\Lambda_k(x)| <r_k\, \forall k\}\subset V$. This proves the first part.
For the second part note that $x_n \to x$ weakly and $\Lambda \in X^{*}$ imply $\Lambda(x_n) \to \Lambda(x)$ because $\Lambda$ is continuous for the weak topology. Conversely, if $\Lambda(x_n) \to \Lambda(x)$ for all $\Lambda \in X^{*}$ and $V$ is any neighborhood of $x$ then $V-x$ is a neigborhood of $0$ and so it contains a set of the form $\cap_k \{x:|\Lambda_k(x)| <r_k\, \forall k\}$. Since $\Lambda_k(x_n) \to \Lambda_k(x)$ for each $k$ we see that $|\Lambda_k(x_n)-\Lambda_k(x)| <r_k$ for each $k$ provided $n$ is sufficiently large. Now you see that $x_N \in V$ for $n$ is sufficiently large.