I am studying the proof of the Banach–Alaoglu theorem in V. Moretti's Spectral Theory and Quantum Mechanics, and it seems an important step in the proof was omitted. (In fact, I think the omitted step is significant enough to render the proof incomplete without it.)
Let $X$ be a normed space over $\mathbb C$, $X'$ its topological dual (the space of bounded linear functionals on $X$), and define the following sets:
$B_x=\{c\in\mathbb C:|c|\leqslant\|x\|\}$
$P=\prod_{x\in X}B_x$, equipped with the product topology
$B=\{f\in X':\|f\|\leqslant1\}$
The theorem states that $B$ is compact in the *-weak topology on $X'$, and part of the proof relies on $B$ being closed in the product topology on $P$. But the proof only shows sequential closure of $B$, that is, that $B\ni p_n\to p\in P$ as $n\to\infty$ in the product topology on $P$ implies $p\in B$. But not all sequentially closed sets are closed, except in the special case of sequential spaces.
Unfortunately, there is even an additional lapse in the proof of sequential closure. The proof claims $p$ is linear because, for any $a, b \in \mathbb C$, $x, y \in X:$
$$p(ax+by)=\lim_{n\to\infty}p_n(ax+by) = a \lim_{n\to\infty}p_n(x) + b\lim_{n\to\infty}p_n(y)=ap(x) + bp(y).$$
The second $=$ sign above is justified only if we already know that $p_n(x)$ and $p_n(y)$ converge for any $x,y \in X$, that is, if $(p_n)$ converges in the *-weak topology. But we are only given that $(p_n)$ converges in the product topology on $P$.
How can these two flaws be remedied? In particular,
1. How can we show that $P$ with the product topology is sequential?
2. How can we justify the second $=$ sign mentioned above?
I would be fine with, instead of answers to (1) and (2), taking a different strategy to correct the flaws mentioned. However, please avoid using nets, as I am not familiar with them and am curious to see how this proof can be accomplished without them.
$B$ is closed because its complement in $X'$ is the union of $\{f\in X': |f(x)| >1\}$ over all vectors $x$ with $\|x\| \leq 1$. The set $\{f\in X': |f(x)| >1\}$ is open in the product topology restricted to $X'$ and union of open sets is open. The fact that $X'$ is closed in $P$ follows from the argument below. Hence the complement in $P$ is also open.
For the second question here is a hint: $\{p\in P: p(ax+by)-ap(x)-bp(y)\neq 0\}$ is open in the product topology.