Daniel Fischer said it all. I would like to make a precision which would have been a comment if I had enough reputation.
The key point is that the closed unit ball of $X^*$ is compact for the weak*-topology. That's called Banach-Alaoglu theorem.
Now to get your equivalence, you also need the following easier facts:
the weak*-topology is Hausdorff.
a compact in a Hausdorff space is closed (and Hausdorff is needed).
a closed subset of a compact space is always compact.
a norm bounded set in a normed vector space is contained in a positive-scalar-multiple of the closed unit ball.
to show that weak*-compact implies norm bounded, you need the Uniform Boundedness Principle and the fact that the image of a compact space by a continuous complex-valued map (point evaluations in this case, which are continuous by definition of the weak*-topology) is compact in $\mathbb{C}$, whence bounded.
Compactness does not imply sequential compactness.
Compactness implies that every sequence has an accumulation point, which is equivalent to countable compactness [every countable open cover has a finite subcover]. But in general, a sequence having accumulation points does not imply that the sequence has a convergent subsequence. One needs additional hypotheses, e.g. first countability of the space to have that implication.
One example of a space that is compact but not sequentially compact is, as shown by the example, the closed unit ball of $(\ell^\infty)^\ast$ in the weak$^\ast$ topology.
A perhaps easier to visualize example is a product of sufficiently many copies of $\{0,1\}$. (Any example must be somewhat difficult to visualize, since the easy-to-visualize spaces have a strong tendency to be first-countable.)
Let $\mathscr{P}(\mathbb{N})$ denote the power set of $\mathbb{N}$, and $X = \{0,1\}^{\mathscr{P}(\mathbb{N})}$ (that is up to the naming of the indices $\{0,1\}^{\mathbb{R}}$, but taking $\mathscr{P}(\mathbb{N})$ makes it easier to define a sequence without convergent subsequences). Define the sequence $(x_n)_{n\in\mathbb{N}}$ in $X$ by
$$p_M(x_n) = \begin{cases} 0 &, n \notin M\\
0 &, n\in M \text{ and } \operatorname{card} \{m\in M : m < n\} \text{ even}\\
1 &, n\in M \text{ and } \operatorname{card} \{ m\in M : m < n\} \text{ odd},\end{cases}$$
where $p_M \colon X \to \{0,1\}$ is the coordinate projection. Then $(x_n)_{n\in\mathbb{N}}$ has no convergent subsequences. For if $(x_{n_k})_{k\in\mathbb{N}}$ is a subsequence, consider the set $M = \{ n_k : k\in\mathbb{N}\}$. Then $p_M(x_{n_k})$ is $0$ for even $k$ and $1$ for odd $k$ (if you follow the convention $0\notin \mathbb{N}$, switch even and odd), so $(x_{n_k})$ is not convergent.
If $E$ is a normed space, then the closed unit ball of $E^\ast$ is compact in the weak$^\ast$ topology by the Banach-Alaoglu theorem, and under some conditions on $E$ it is also sequentially compact.
- If $E$ is separable, then the subspace topology induced on the closed unit ball of $E^\ast$ by the weak$^\ast$ topology is metrisable (Note: The weak$^\ast$ topology on $E^\ast$ is then generally not metrisable itself), hence the closed unit ball of $E^\ast$ is then weak$^\ast$-sequentially compact.
- If $E$ is reflexive, the closed unit ball of $E^\ast$ is weak$^\ast$-sequentially compact.
$\ell^\infty$ is neither separable nor reflexive.
Best Answer
Let me answer your second question first.
The weak$^{\ast}$-topology is Hausdorff (let me treat the real case, the complex case is similar): If $\phi \neq \psi$ are two linear functionals then there is $x \in X$ such that $\phi(x) \lt r \lt \psi(x)$. The sets $U = \{f \in X^{\ast} \,:\,f(x) \lt r\}$ and $V = \{f \in X^{\ast}\,:\,f(x) \gt r\}$ are weak$^{\ast}$-open (since evaluation at $x$ is weak$^{\ast}$-continuous) and disjoint neighborhoods of $\phi$ and $\psi$, respectively.
That the weak topology is Hausdorff is shown similarly, using Hahn-Banach.
Next, if $X$ is separable, then the unit ball in the dual space is metrizable with respect to the weak$^{\ast}$-topology: pick a countable dense set $\{x_{n}\}_{n \in \mathbb{N}}$ of the unit ball of $X$ and verify that \[ d(\phi,\psi) = \sum_{n=1}^{\infty} 2^{-n} \frac{|\phi(x_n) - \psi(x_n)|}{1+|\phi(x_n) - \psi(x_n)|} \] defines a metric compatible with the weak$^{\ast}$-topology. Hence the unit ball is sequentially compact in the weak$^{\ast}$-topology (this can be shown directly using Arzelà-Ascoli, by the way).
Using a standard Baire category argument, one can show that weak$^{\ast}$-compact sets are norm-bounded: Indeed, if $K$ is weak$^{\ast}$-compact, it is a Baire space. Write $B^{\ast}$ for the closed unit ball in $X^{\ast}$. Clearly $K = \bigcup_{n = 1}^{\infty} (K \cap n \cdot B^{\ast})$, so at least one of the closed subsets $K \cap n \cdot B^{\ast}$ of $K$ must have non-empty interior. By compactness finitely many translates of $n\cdot B^{\ast}$ must cover $K$, thus $K$ is bounded in norm and hence $K$ is a closed subset of a large enough ball.
I don't know if the converse is true.
If $X$ is not separable, then weak$^{\ast}$-compactness does not imply weak$^{\ast}$-sequential compactness, the standard example is mentioned in Florian's post.
Since you might be interested in the weak topology as well, there's a rather difficult result due to Eberlein:
Recall that a space is countably compact if every countable open cover has a finite subcover. A sequentially compact space is countably compact.
and finally: