Banach-Alaoglu Theorem

Theorem
The closed unit ball of the dual of a normed space is compact with respect to the weak* topology.

Proof
Let $X$ be a normed vector space, denote by $B$ the closed unit ball in $X$. Let $X^*$ be the dual of $X$ and denote by $B^*$ the closed unit ball in $X^*$. Let $\mathcal F(B) = [-1,1]^B$ be the topological space of functions from $B$ to $[-1,1]$. By Tychonoff's Theorem, $\mathcal F(B)$ is compact with respect to the product topology.

We define the restriction map $R: B^*\rightarrow \mathcal F(B)$ by $R(\psi) = \psi|_B$.

Lemma 1
$R(B^*)$ is a closed subset of $\mathcal F(B)$.

Lemma 2
$R$ is a homeomorphism from $B^*$ with the weak* topology to its image $R(B^*)$ seen as a subset of $\mathcal F(B)$ with the product topology.

Proof
Thus by lemma 2, $B^*$ in the weak* topology is homeomorphic with $R(B^*)$, which is a closed subset of $\mathcal F(B)$ (by lemma 1) and thus compact.