I'd like to check that I understand the proof in full detail. Can you tell me if the following is correct? Thanks for your help.

Claim: The closed unit ball $B_{\|\cdot\|_{op}}(0,1)$ in $X^\ast$ is compact in the weak$^\ast$ topology.

Proof: Define $Y = \prod_{x \in X} B(0, \|x\|)$, the product of closed balls of radius $\|x\|$ in $\mathbb R$ (or $\mathbb C$). Then by Tychonoff's theorem, $Y$ is compact.

Now define a map $\phi : B_{\|\cdot\|_{op}}(0,1) \to Y$ by $\ell \mapsto (\ell(x))_x$. Then the image $\mathrm{im} \phi$ are sequences with $x$-th entry $\|\ell(x)\| \leq \|x\|$. Hence *if* the things in the image are linear, then they are in the image of $\phi$. So everything outside the image, that is, lying in $Y \setminus \mathrm{im} \phi$ is non-linear. (That all of $Y$ are functionals, that is, maps $X \to \mathbb C$, is clear by definition).

So we have established that the image $\phi (B_{\|\cdot\|_{op}}(0,1))$ is $B_{\|\cdot\|_{op}}(0,1)$ itself. Hence $\varphi$ is surjective. That it is injective is clear. We proceed by showing that it's a homeomorphism and finish the proof by showing that its image is closed:

First, $\phi$ is a homeomorphism: We already know that $\phi$ is continuous, by definition (we put the weak$^\ast$ topology on its domain and $\phi$ is an evaluation map). Its inverse is also continuous: Let $y \in B_{\|\cdot\|_{op}}(0,1)$ be a point and $\prod_{x \in X} B(y(x),\varepsilon_x) = \prod_{x \in X} B(\ell(x),\varepsilon_x)$ an open set containing it. Then since we have the product topology, all but finitely many components are the entire space, that is, for all but finitely many $x$ we have $B(\ell(x),\varepsilon_x) = B(0, \|x\|)$. So that $\phi^{-1}$ of an open set like this is a finite intersection: $$ \phi^{-1}(\prod_{x \in X} B(\ell(x),\varepsilon_x)) = \bigcap_{i=1}^n \{ \ell^\prime \in B_{\|\cdot\|_{op}}(0,1) \mid |\ell (x) - \ell^\prime (x)| < \varepsilon_x \} \supset \bigcap_{i=1}^n \{ \ell^\prime \in B_{\|\cdot\|_{op}}(0,1) \mid |\ell (x) - \ell^\prime (x)| < \varepsilon \}$$ where $\varepsilon = \min(\varepsilon_x)$. So that $\phi^{-1}$ of an open set maps to a basic open set in $B_{\|\cdot\|_{op}}(0,1)$.

To finish we need to show that $\phi (B_{\|\cdot\|_{op}}(0,1))$ is closed: We show instead that its complement is open. To this end, let $y \in Y \setminus \phi (B_{\|\cdot\|_{op}}(0,1))$. Then $y$ is a non-linear functional so that there exist scalars $a,b$ and $x_1, x_2 \in X$ such that $y(ax_1 + bx_2) \neq a y(x_1) + b y(x_2)$. Choose $\varepsilon$ such that $B(y(ax_1 + bx_2), \varepsilon) \cap (a B(y(x_1), \varepsilon) + b B(y(x_2), \varepsilon)) = \varnothing$. If $\pi_x: Y \to B(0, \|x\|)$ is the projection then $\pi_x$ is continuous and hence its inverse maps open sets to open sets so that it's enough to show that the intersection of the inverse images of the open balls lies entirely in the complement of the image of $\varphi$: Let $z \in \pi_{ax_1 + bx_2}^{-1}(B(y(ax_1 + bx_2), \varepsilon)) \cap (\pi_{ax_1}^{-1} aB(y(x_1), \varepsilon) \cap \pi_{ax_1}^{-1} bB(y(x_2), \varepsilon)$. Then $z(ax_1 + bx_2) \neq a z(x_1) + b z(x_2)$.