Banach-Alaoglu Theorem

Theorem
Let $X$ be a separable normed space. Then the closed unit sphere in its dual $X^*$ is weak* sequentially compact.

Proof
We have to show the follwing: given a bounded sequence in $X^*$, there is a weakly convergent subsequence.

Let $\{l_n\}_{n\in\N}$ be a bounded sequence in $X^*$. Let $\{x_n\}_{n\in\N}$ be a countable dense subset of $X$.

Choose subsequences $\N\supset\Lambda_1\supset\Lambda_2\supset\ldots$ such that for every $j\in\N$ we have $l_k(x_j)\rightarrow a_j =: l(x_j)$ as $k\rightarrow\infty$, $k\in\Lambda_j$.

let $\Lambda$ be the diagonal sequence.

Claim 1
$l$ can be extended to an element of $X^*$.

Proof
$l$ can be extended in the obvious way to a linear function on $M = \mathbf{span}\{x_j\}_{j\in\N}$. We extend it to a functional in $X^*$ by pointwise limit (notice that $M$ is dense in $X$). We have: $$|l(x)| = \lim_{k\rightarrow\infty}|l(x_k)|\leq\limsup_{k\rightarrow\infty}||l_k||_{X^*}||x||_X$$ where $x_k\rightarrow x$ as $k\rightarrow\infty$. Since $\{l_k\}_k\in\N$ was bounded, $l$ is bounded and thus continuous.

Claim 2
$l_k\stackrel{\omega^*}{\rightarrow}l$ as $k\rightarrow\infty$, $k\in\Lambda$.

Proof
Let $X\ni x = \lim_{j\rightarrow\infty,\ j\in J}x_j$, where $J$ is some subset of $\N$. We have then: $$|l_k(x) - l(x)|\leq |l_k(x - x_j)| + |l(x - x_j)| + |l_k(x_j) - l(x_j)|\leq$$ $$\leq (\sup_{i\in\Lambda}||l_i||_{X^*} + ||l||_{X^*})||x - x_j||_X + |l_k(x_j) - l(x_j)|\stackrel{k,j\rightarrow\infty}{\rightarrow}0$$