I'm trying to understand this passage from Rudin's Functional Analysis:
Since every weak neighborhoud of $0$ contains a neighborhood of the form $V = \{x: |T_ix| < r_i, 1 \le i \le n\}$ where $T_i \in X^*$ and $r_i>0$ ...
Does this mean that for every weak neighborhoud $U$ of $0$ there exists $T_1 \ldots, T_n \in X^*$ and $r_1, \ldots, r_n$ positive such that $U \supset \{x: |T_ix| < r_i, 1 \le i \le n\} $?
He goes on...
This implies that $x_n \to 0$ weakly $\iff$ $Tx_n \to 0$ for all $T$ in $X^*$.
How does this follow?
For your first question, yes you interpreted it correctly. For the second, suppose $x_n\to 0$ weakly and $T\in X^*$. Given $\varepsilon>0$, the set $U=\{x\in X:|Tx|<\varepsilon\}$ is a weak open neighborhood of $0$, hence there is some $N\in\mathbb N$ such that $x_n\in U$ for all $n\geq N$. In other words, $|Tx_n|<\varepsilon$ for all $n\geq N$. Since $\varepsilon>0$ was arbitrary, we have $Tx_n\to 0$.
For the reverse direction, let $U$ be a weak open neighborhood of $0$. Then there exist $T_1,\ldots,T_m\in X^*$ and $r_1,\ldots,r_m>0$ such that $\{x: |T_ix| < r_i, 1 \le i \le m\}\subset U$. For $1\leq i\leq m$. there is some $N_i\in\mathbb N$ such that $|T_ix_n|<r_i$ for $n\geq N_i$. Let $N=\max\{N_i:1\leq i\leq m\}$. Then $|T_ix_n|<_i$ for all $i$ whenever $n\geq N$. That is, $x_n\in U$ for all $n\geq N$. Since the neighborhood $U$ was arbitrary, we conclude that $x_n\to0$ weakly.