[proofplan] We embed the closed unit ball $B_{X^*}$ of the dual space into a product of compact intervals via the evaluation map $\Phi$, invoke Tychonoff's theorem to make the ambient product space compact, show that $\Phi$ is a homeomorphism onto its image when both spaces carry the weak-* and product topologies respectively, and prove that the image $\Phi(B_{X^*})$ is closed in the product space. Since a closed subset of a compact space is compact, the result follows. [/proofplan] [step:Embed $B_{X^*}$ into a compact product space via the evaluation map] For each $x \in X$, define the compact interval \begin{align*} D_x := \{z \in \mathbb{R} : |z| \leq \|x\|_X\} = [-\|x\|_X, \, \|x\|_X]. \end{align*} Each $D_x$ is a closed bounded subset of $\mathbb{R}$, hence compact by the Heine-Borel theorem. Form the product space \begin{align*} P := \prod_{x \in X} D_x, \end{align*} equipped with the product topology. By Tychonoff's theorem, $P$ is compact. Define the evaluation map \begin{align*} \Phi: B_{X^*} &\to P \\ f &\mapsto (f(x))_{x \in X}. \end{align*} For any $f \in B_{X^*}$, the operator norm condition $\|f\|_{X^*} \leq 1$ gives $|f(x)| \leq \|f\|_{X^*} \|x\|_X \leq \|x\|_X$ for every $x \in X$, so $f(x) \in D_x$ and $\Phi(f) \in P$. The map $\Phi$ is well-defined. [guided] The idea is to realise $B_{X^*}$ as a subset of a space that is already known to be compact. For each $x \in X$, the compact interval $D_x = [-\|x\|_X, \|x\|_X]$ is the range of the $x$-th coordinate of $\Phi(f)$, namely $f(x)$. The product \begin{align*} P = \prod_{x \in X} D_x \end{align*} is an enormous space -- one coordinate for each vector in $X$ -- but Tychonoff's theorem guarantees compactness of any product of compact spaces, regardless of the cardinality of the index set $X$. Why does the operator norm condition $\|f\|_{X^*} \leq 1$ matter here? For any $f \in B_{X^*}$ and any $x \in X$, the definition of the operator norm gives $|f(x)| \leq \|f\|_{X^*}\|x\|_X \leq \|x\|_X$, so $f(x) \in D_x$. This confines $\Phi(f) = (f(x))_{x \in X}$ to the product $P$. Without this uniform bound on the functionals, the image of $\Phi$ would not be contained in a product of compact sets, and Tychonoff would not apply. The evaluation map $\Phi: B_{X^*} \to P$ defined by $\Phi(f) = (f(x))_{x \in X}$ is therefore well-defined. [/guided] [/step] [step:Show $\Phi$ is a homeomorphism from $(B_{X^*}, \text{weak-*})$ onto its image in $(P, \text{product})$] The weak-* topology on $X^*$ is defined as the coarsest topology making all evaluation maps $\mathrm{ev}_x: f \mapsto f(x)$ continuous for each $x \in X$. The product topology on $P$ is the coarsest topology making all coordinate projections $\pi_x: (z_y)_{y \in X} \mapsto z_x$ continuous. Since $\pi_x \circ \Phi = \mathrm{ev}_x$ for each $x \in X$, the map $\Phi$ is continuous from $(B_{X^*}, \text{weak-*})$ to $(P, \text{product})$. The map $\Phi$ is injective: if $\Phi(f) = \Phi(g)$, then $f(x) = g(x)$ for all $x \in X$, so $f = g$ as elements of $X^*$. The inverse $\Phi^{-1}: \Phi(B_{X^*}) \to B_{X^*}$ is continuous because the weak-* topology on $B_{X^*}$ is the initial topology with respect to the maps $\mathrm{ev}_x$, and each $\mathrm{ev}_x \circ \Phi^{-1} = \pi_x|_{\Phi(B_{X^*})}$ is continuous. Therefore $\Phi$ is a homeomorphism onto its image. [/step] [step:Prove $\Phi(B_{X^*})$ is a closed subset of $P$] Let $(g_\alpha)$ be a net in $\Phi(B_{X^*})$ converging to some $g = (g(x))_{x \in X} \in P$ in the product topology. For each $\alpha$, there exists $f_\alpha \in B_{X^*}$ with $g_\alpha = \Phi(f_\alpha)$, meaning $g_\alpha(x) = f_\alpha(x)$ for every $x \in X$. Product-topology convergence means $f_\alpha(x) \to g(x)$ for every $x \in X$. We verify that $g$ defines a bounded linear functional on $X$ with $\|g\|_{X^*} \leq 1$. **Linearity:** For any $x, y \in X$ and $\alpha_0, \beta_0 \in \mathbb{R}$, each $f_\alpha$ is linear, so $f_\alpha(\alpha_0 x + \beta_0 y) = \alpha_0 f_\alpha(x) + \beta_0 f_\alpha(y)$. Taking the limit: $g(\alpha_0 x + \beta_0 y) = \alpha_0 g(x) + \beta_0 g(y)$. **Boundedness:** Since $g \in P$, we have $|g(x)| \leq \|x\|_X$ for every $x \in X$ (this is the constraint defining $D_x$). Therefore $\|g\|_{X^*} = \sup_{\|x\|_X \leq 1} |g(x)| \leq 1$, so $g \in B_{X^*}$. Hence $g = \Phi(g) \in \Phi(B_{X^*})$, and $\Phi(B_{X^*})$ is closed in $P$. [guided] The closedness argument is the heart of the proof. We must show that any limit point of $\Phi(B_{X^*})$ in the product topology still represents a bounded linear functional. Let $(g_\alpha)$ be a net in $\Phi(B_{X^*})$ converging to $g \in P$. For each $\alpha$, write $g_\alpha = \Phi(f_\alpha)$ with $f_\alpha \in B_{X^*}$. **Linearity is preserved** because it is an identity involving finitely many coordinates at a time. For any $x, y \in X$ and scalars $\alpha_0, \beta_0$: \begin{align*} g(\alpha_0 x + \beta_0 y) = \lim_\alpha f_\alpha(\alpha_0 x + \beta_0 y) = \lim_\alpha (\alpha_0 f_\alpha(x) + \beta_0 f_\alpha(y)) = \alpha_0 g(x) + \beta_0 g(y). \end{align*} **The norm bound is preserved** because each $D_x = [-\|x\|_X, \|x\|_X]$ is a closed interval. Since $g \in P$, we have $|g(x)| \leq \|x\|_X$ for every $x \in X$, giving $\|g\|_{X^*} = \sup_{\|x\|_X \leq 1}|g(x)| \leq 1$. So $g \in B_{X^*}$ and $g = \Phi(g) \in \Phi(B_{X^*})$. What would fail if we tried the same argument with the norm topology on $X^*$? The set $B_{X^*}$ is closed and bounded in the norm topology, but in infinite dimensions, closed and bounded does not imply compact (the closed unit ball of an infinite-dimensional normed space is never norm-compact). The weak-* topology is strictly coarser, which makes it easier for a set to be compact -- there are fewer open covers to worry about. [/guided] [/step] [step:Conclude weak-* compactness of $B_{X^*}$] The product space $P$ is compact by Tychonoff's theorem (from the first step), and $\Phi(B_{X^*})$ is a closed subset of $P$ (from the previous step). A closed subset of a compact space is compact, so $\Phi(B_{X^*})$ is compact in the product topology. Since $\Phi$ is a homeomorphism from $(B_{X^*}, \text{weak-*})$ onto $(\Phi(B_{X^*}), \text{product})$, the preimage $B_{X^*}$ is compact in the weak-* topology. [/step]