Let $E$ be a normed space and $E^*$ its topological dual. The weak topology $\sigma(E, E^*)$ is the initial topology on $E$ w.r.t. the collection $E^*$ of maps. Let $(x_d)_{d \in D}$ be a net in $E$ and $x \in E$.
I'm trying to adapt the proof for sequence to that for net. Could you have a check on my attempt?
Theorem: $x_d \to x$ in $\sigma(E, E^*)$ if and only if $\langle f, x_d \rangle \to \langle f, x \rangle$ for all $f \in E^*$.
Proof: One direction is obvious. Let's prove the converse. Let $U$ be a neighborhood of $x$ in $\sigma(E, E^*)$. Then there are $f_1, \ldots, f_n \in E^*$ and neighborhoods $V_1, \ldots, V_n$ of $\langle f_1, x \rangle, \ldots, \langle f_n, x \rangle$ respectively such that $$ \bigcap_{i=1}^n f_i^{-1} (V_i) \subset U. $$
Because $\langle f_i, x_d \rangle \to \langle f_i, x \rangle$, there is $d_i$ such that $\langle f_i, x_d \rangle \in V_i$ for all $d$ such that $d_i \le d$. Let $\overline d$ be the upper bound of $\{d_1, \ldots, d_n\}$. Then $$ \langle f_i, x_d \rangle \in V_i \quad \forall \overline d \le d, \forall i = 1, \ldots,n. $$ and thus $$ x_d \in f_i^{-1}(V_i) \quad \forall \overline d \le d, \forall i = 1, \ldots,n. $$
It follows that $$ x_d \in \bigcap_{i=1}^n f_i^{-1} (V_i) \quad \forall \overline d \le d. $$
Hence $$ x_d \in U \quad \forall \overline d \le d. $$
This implies $x_d \to x$. This completes the proof.
Remark: The weak$^*$ topology $\sigma(E^*, E)$ is the initial topology on $E^*$ w.r.t. the collection $(\varphi_x)_{x\in E}$ of maps $$ \varphi_x: E^* \to \mathbb R, f \mapsto \langle f, x \rangle. $$
Let $(f_d)_{d \in D}$ be a net in $E^*$ and $f \in E^*$. With a similar reasoning as above, we have
Theorem: $f_d \to f$ in $\sigma(E^*, E)$ if and only if $\langle f_d, x \rangle \to \langle f, x \rangle$ for all $x \in E$.
Update: Above result can be generalized to easily to arbitrary initial topology as below.
Given a set $X$ and an indexed family $(Y_{i}, \tau_i)_{i \in I}$ of topological spaces with functions $f_{i}: X \rightarrow Y_{i}$. The initial topology $\tau$ on $X$ induced by the collection $(f_i)_{i\in I}$ is the coarsest topology on $X$ such that each $$ f_{i}:(X, \tau) \to (Y_{i}, \tau_i) $$ is continuous. It follows that $\{f_i^{-1} (O_i) \mid i\in I, O_i \in \tau_i\}$ is a subbase of $\tau$.
Theorem 1: If $(x_d)$ is a net in $X$ and $x\in X$. Then $x_d \to x$ in $\tau$ if and only if $f_i(x_d) \to f_i (x)$ for all $i \in I$.
Proof: One direction is obvious. Let's prove the converse. Let $U$ be a neighborhood of $x$ in $\tau$. Then there are $f_1, \ldots, f_n$ and neighborhoods $V_1, \ldots, V_n$ of $f_1 (x), \ldots, f_n (x)$ in $\tau_1, \ldots, \tau_n$ respectively such that $$ \bigcap_{i=1}^n f_i^{-1} (V_i) \subset U. $$
Because $f_i (x_d) \to f_i (x)$, there is $d_i$ such that $f_i (x_d) \in V_i$ for all $d$ such that $d_i \le d$. Let $\overline d$ be an upper bound of $\{d_1, \ldots, d_n\}$. Then $$ f_i (x_d) \in V_i \quad \forall \overline d \le d, \forall i = 1, \ldots,n. $$ and thus $$ x_d \in f_i^{-1}(V_i) \quad \forall \overline d \le d, \forall i = 1, \ldots,n. $$
It follows that $$ x_d \in \bigcap_{i=1}^n f_i^{-1} (V_i) \quad \forall \overline d \le d. $$
Hence $$ x_d \in U \quad \forall \overline d \le d. $$
This implies $x_d \to x$. This completes the proof.
Theorem 2: Let $Z$ be a topological space and $g:Z \to X$. Then $g$ is continuous if and only if $f_i \circ g$ is continuous for all $i \in I$.
Proof: One direction is obvious. Let's prove the converse. Let $(z_d)$ be a net in $Z$ such that $z_d \to z \in Z$. Let $z'_d := g(z_d) \in X$ and $z' := g(z) \in X$. By continuity of $f_i \circ g$, we have $f_i \circ g (z_d) \to f_i \circ g (z)$ and thus $f_i (z'_d) \to f_i (z')$ for all $i\in I$. By Theorem 1, we get $z'_d \to z'$ in $\tau$. Hence $g$ is continuous.