= Solution
The <Tychonoff theorem> states that an arbitrary product of <compact spaces>, with the <product topology>, is compact; no <Hausdorff> hypothesis is necessary. We prove it using <Zorn's lemma> and the <axiom of choice>.
First, every proper <filter on a set> extends to an <ultrafilter>. Order its proper extensions by inclusion. The union of a chain is again a proper filter: any finite collection of its members occurs in one member of the chain, and the empty set occurs in none. <Zorn's lemma> gives a maximal proper filter $\mathcal U$. If $A\notin\mathcal U$, adjoining $A$ must destroy properness, so some $F\in\mathcal U$ has $F\cap A=\varnothing$, and hence $X\setminus A\in\mathcal U$. This proves the decision property of an <ultrafilter>.
Next prove the <ultrafilter characterization of compactness>. In a <compact space> the <closed sets> $\overline F$, for $F\in\mathcal U$, have the <finite intersection property>, so their intersection contains a point $x$. Every open neighbourhood $O$ of $x$ belongs to $\mathcal U$: otherwise its closed complement belongs to $\mathcal U$ and contains $x$, a contradiction. Thus the ultrafilter converges to $x$. Conversely, if an open cover has no finite subcover, the closed complements have the <finite intersection property> and generate a proper filter. Extend it to an ultrafilter. A limit point lies in some member $O$ of the cover, so both $O$ and its complement would belong to the ultrafilter. This is impossible. These arguments also prove the closed-set form of <compactness> directly by taking complements.
Now let $X=\prod_{j\in J}X_j$. If a factor is empty, the product is empty and compact. Otherwise the <axiom of choice> makes $X$ nonempty. For an <ultrafilter> $\mathcal U$ on $X$, its coordinate image
$$
\mathcal U_j=\{A\subseteq X_j:\pi_j^{-1}(A)\in\mathcal U\}
$$
is an ultrafilter. <Compactness> of $X_j$ supplies at least one limit $x_j$; use the <axiom of choice> to choose one for every $j$. A basic open neighbourhood of $x=(x_j)_j$ restricts finitely many coordinates. Each corresponding inverse image belongs to $\mathcal U$, so their finite intersection also does. Hence $\mathcal U$ converges to $x$ in the <product topology>. The ultrafilter criterion proves the theorem. The explicit choices and the maximal-filter argument are exactly the places where choice enters.
The <Banach-Alaoglu theorem> states that the closed <unit ball> of the <continuous dual space> $V^*$ of a real or complex <normed vector space> $V$ is compact in the <weak-star topology> $\sigma(V^*,V)$. Completeness of $V$ is not required.
For each $v\in V$, let $D_v=\{z\in\mathbb K:|z|\leq\|v\|\}$, a compact scalar disk or interval. By the <Tychonoff theorem>, $P=\prod_{v\in V}D_v$ is compact. Inside $P$, impose the equations
$$
z_{u+v}=z_u+z_v,\qquad z_{\alpha v}=\alpha z_v
\quad(u,v\in V,\ \alpha\in\mathbb K).
$$
Each equation defines a closed subset, since it concerns continuous coordinate projections into a <Hausdorff> scalar <field>. Their intersection $K$ is therefore compact. A point in $K$ defines a <linear functional> $\ell(v)=z_v$, and the coordinate bounds give $|\ell(v)|\leq\|v\|$. Conversely, every $\ell\in V^*$ with $\|\ell\|\leq1$ gives such a point. Thus $K$ is exactly the image of the dual <unit ball> under $\ell\mapsto(\ell(v))_{v\in V}$. The induced <product topology> is precisely pointwise convergence on $V$, which is the <weak-star topology>. This proves
$$
\boxed{\{\ell\in V^*:\|\ell\|\leq1\}\text{ is weak-star compact}.}
$$
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