The weak$^\star$ topology $\sigma(E',E)$ coincides with the subspace topology that $\tau$ induces $E'$

Question

I'm reading an elegant proof Banach–Alaoglu theorem from here. The proof depends on the following observation.

Let $E$ be a normed vector space, $E'$ its topological dual, and $\sigma(E', E)$ the weak$^\star$ topology of $E'$. We endow $X := \mathbb R^E$ with the product topology denoted by $\tau$. Clearly, $E' \subseteq X$. Then $\sigma(E',E)$ coincides with the subspace topology that $\tau$ induces $E'$.

Could you have a check on my attempt?

PS: I posted my proof separately so that I can accept my own answer to remove my question from unanswered list. Surely, if other people post answers, then I will happily accept theirs.

Answer 1

For $x\in E$, let $\pi_x:X \to\mathbb R, f \mapsto f(x)$ and $\varphi_x: E' \to \mathbb R, f \mapsto f(x)$. Then $$ \mathcal B:= \{ \pi_{x} ^{-1} [A] \mid x\in E, A \text{ is an open subset of } \mathbb R\} $$ is a subbase of $\tau$. Also, $$ \mathcal D:= \{ \varphi_{x} ^{-1} [A] \mid x\in E, A \text{ is an open subset of } \mathbb R\} $$ is a subbase of $\sigma(E',E)$. We have $$ \begin{aligned} f \in \varphi_{x} ^{-1} [A] &\iff f \in E' \text{ and } \varphi_x(f) \in A \\ &\iff f\in E' \text{ and } f(x) \in A \\ &\iff f\in E' \text{ and } \pi_x(f) \in A \\ &\iff f\in E' \text{ and } f \in \pi_{x} ^{-1} [A] \\ &\iff f\in E' \cap \pi_{x} ^{-1} [A]. \end{aligned} $$

It follows that $\mathcal D = \{E' \cap B \mid B \in \mathcal B\}$. This completes the proof.