= Solution
The topology needed here is the <weak-star topology> $\sigma(E^*,E)$, meaning pointwise convergence on $E$. With the literal <weak topology> $\sigma(E^*,E^{**})$, the claimed <compactness> is false. For example, $E=c_0$ is separable and $E^*=\ell^1$. Its unit <vectors> $e_n$ have no weak cluster point: every coordinate functional forces a cluster point to have all coordinates zero, while the bounded functional $x\mapsto\sum_jx_j$ has value one throughout. Thus the dual <unit ball> is not <weakly compact>. We prove the intended weak-star assertion in full.
Choose a countable dense set $(x_k)$ in the <unit ball> of $E$. On $E_1^*$ define
$$
\boxed{d(\phi,\psi)=\sum_{k\geq1}2^{-k}|\phi(x_k)-\psi(x_k)|.}
$$
Each term is at most $2\cdot2^{-k}$, so the series converges. The <triangle inequality> is immediate, and $d=0$ forces equality on the dense set and hence on all of $E$. It is therefore a metric. Convergence in this metric is equivalent to convergence on every $x_k$: one direction follows from any individual term, and the other by bounding the series tail uniformly. Because $\|\phi\|,\|\psi\|\leq1$, approximation of any unit <vector> by $x_k$ then gives convergence on all of $E$. The same finite-tail and approximation argument for neighborhoods proves equality with the <weak-star topology>, not just agreement of its convergent sequences.
Given a sequence $(\phi_n)$ in this ball, diagonal selection gives a subsequence converging at every $x_k$, since each coordinate lies in a compact disk. Uniform <norm> bounds then make it pointwise Cauchy at every $x\in E$. Define $\phi(x)$ as the limit. <Linearity> passes to the limit, and $|\phi(x)|\leq\|x\|$, so $\phi\in E_1^*$. The subsequence converges to it in the metric. Thus this <metric space> is sequentially compact, hence compact, proving
$$
\boxed{E_1^*\text{ is compact and metrizable in }\sigma(E^*,E).}
$$
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