I'm proving this identity about the product of weak topologies. Could you have a check on my attempt?
Let $E,F$ be T.V.S. and $E^\star, F^\star$ their continuous dual respectively. Let $\sigma (E, E^\star)$ and $\sigma (F, F^\star)$ be the weak topology of $E, F$ respectively. Let $\sigma (E, E^\star) \boxtimes \sigma (F, F^\star)$ be the product topology of $\sigma (E, E^\star)$ and $\sigma (F, F^\star)$. Then $$\sigma (E, E^\star) \boxtimes \sigma (F, F^\star) = \sigma \big (E \times F, (E \times F)^\star \big).$$
My attempt: Let $B_1, B_2$ be the basis of $\sigma (E, E^\star)$ and $\sigma (F, F^\star)$ respectively. Here $B_1$ is the collection of all sets of the form $\bigcap_{i\in I} f_i^{-1}[O_i]$ where $I$ is finite, $(f_i)_{i\in I}$ a subset of $E^\star$, and $(O_i)_{i \in I}$ a collection of open sets in $\mathbb R$.
Similarly, $B_2$ is the collection of all sets of the form $\bigcap_{j\in J} h_j^{-1}[O_j]$ where $J$ is finite, $(h_j)_{j\in J}$ a subset of $F^\star$, and $(O_j)_{j \in J}$ a collection of open sets in $\mathbb R$.
Then $B_1 \times B_2$ is a basis of $\sigma (E, E^\star) \boxtimes \sigma (F, F^\star)$. It follows that $B_1 \times B_2$ is the collection of all sets of the form $$\left (\bigcap_{i\in I} f_i^{-1}[O_i] \right ) \times \left (\bigcap_{j\in J} h_j^{-1}[O_j] \right ) = \bigcap_{i\in I, j\in J} f_i^{-1}[O_i] \times h_j^{-1}[O_j].$$
Let $B$ be a basis of $\sigma \big (E \times F, (E \times F)^\star \big)$. Here $B$ is a collection of all sets of the form $$\bigcap_{k\in K} g_k^{-1}[O_k],$$ where $K$ is finite, $(g_k)_{k\in K}$ a subset of $(E \times F)^\star$, and $(O_k)_{k \in K}$ a collection of open sets in $\mathbb R$. Next we need a lemma.
Let $E, F$ be T.V.S. and $E^\star, F^\star$ their continuous dual respectively. Let $f \in E^\star, h\in F^\star$. We define $g : E\times E \to \mathbb R, (x,y) \mapsto f(x)+h(y)$. Then $g\in (E\times F)^\star$. Let $A := (a_1, a_2)$ be an open interval of $\mathbb R$ and $$X: = \bigcup \{f^{-1}(B) \times h^{-1}(C) \mid B,C \text{ open in } \mathbb R \text{ s.t. } B+C=A\}.$$ Then $X=g^{-1} (A)$. [A proof is given here]
For each $g \in (E \times F)^\star$, there is a unique pair $(f, h) \in E^\star \times F^\star$ such that $g(x,y)=f(x)+h(y)$. By our lemma, $$g^{-1} (O) = \bigcup \{f^{-1}(O_1) \times h^{-1}(O_2) \mid O_1,O_2 \text{ open in } \mathbb R \text{ s.t. } O_1+O_2=O\}.$$
Notice that given a collection of $(U_\lambda)_{\lambda \in \Lambda}$, below two constructing orders yield the same topology.
First, we take all finite intersections of sets in $\left(U_{\lambda}\right)_{\lambda \in \Lambda}$, and obtain a new family $\Phi$. Next we obtain a new family $\mathscr{F}$ by forming arbitrary unions of elements from $\Phi$.
First, we take all arbitrary unions of sets in $\left(U_{\lambda}\right)_{\lambda \in \Lambda}$, and obtain a new family $\Phi$. Next we obtain a new family $\mathscr{H}$ by forming finite intersections of elements from $\Phi$. Finally, Next we obtain a new family $\mathscr{F}$ by forming arbitrary unions of elements from $\mathscr{H}$.
It follows that $B_1 \times B_2$ and $B$ are the basis of the same topology. This completes the proof.
Update: I have found a mistake in the proof. Because $g_k^{-1}[O_k]$ is a union of sets in $B_1 \times B_2 $ with some form specified by our lemma, so the topology generated by $B$ is coarser than the one generated by $B_1 \times B_2$. I'm unable to prove the converse, i.e., the topology generated by $B$ is finer than the one generated by $B_1 \times B_2$.