Theorem: If $X$ is a normed vector space, the closed unit ball $B^* := \{ f \in X^* \, : \, \|f\| \le 1 \}$ in $X^*$ is compact in the weak$^*$ topology.
Proof: For each $x \in X$, let $D_x := \{ z \in \mathbb{C} \, : \, |z| \le \|x \| \}$ and $D:= \prod_{x \in X} D_x$, which is compact by Tychnoff's Theorem.
The relative topologies that $B^*$ inherits from the product topology on $D$ and the weak $^*$ topology coincide with topology of pointwise convergence, so it suffices to see that $B^*$ is closed...
Why are the two topologies the same? This is how I would argue:
$D$ has the product topology, and $B^*$ has as a subspace, a topology generated by $$ \{ B^* \cap \pi_x^{-1}(U_k) \, : \, U_k \text{ open in } D_x \}_{x \in X}$$ As a subspace of the weak$^*$ topology, $B^*$ also has another topology given by basis, \begin{align*} \{ B^* \cap x^{-1}(U_k) \, : \, U_k \text{ open in } \mathbb{C} \}_{x \in X} \end{align*} where $x \in X^{**}$ is an identification of $x \in X$. These two sets generate the same topology as $f \in B^* \Rightarrow f(x) \in D_x$, so $B^* \cap x^{-1}(U_k) = B^* \cap x^{-1}(U_k \cap D_x) = B^* \cap \pi_x^{-1}(\hat{U}_k)$ where $\hat{U}_k$ is an open set in $D_x$.
Because a closed subset of a compact set is compact.