Suppose that the unit sphere $S_X$ of $(X, \| . \|_X)$ is compact.
Then the unit ball of $X$ is the image of the compact set $[0,1] \times S_X$ by the continuous map $(t, v) \mapsto tv$, and hence is compact.
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
Idea for an indirect proof: Suposse unit ball is compact. Cover it by $\cup_{x\in B_1 (0)} B_{\frac{1}{2}} (x) $. By compactness, there exist points finitely many points $a_i, i=1,\dots,n$ such that balls of radius $\frac{1}2$ cover unit ball. Then it should be that your space is infact closure of linear span of $a_i$.