My reasoning so far is as follows: Let $E$ be a separable Banach space. Since ${\mathrm{id}:(E,\|\cdot\|)\longrightarrow(E,w)}$ is a continuous surjection and $(E,\|\cdot\|)$ is separable, so too is $(E,w)$. Since ${{}^{\ast\ast}:(E,w)\longrightarrow (E^{\ast\ast},w*)}$ is a bicontinuous bijection (ie an homeomorphism), it follows that $(E^{\ast\ast},w^{\ast})$ is w*-separable. By w*-density of $E^{\ast\ast}$ in $E^{\prime\prime}$ (Alaoglu's Theorem), one has that $E^{\prime\prime}$ is w*-separable.
Remark. The corollary that I need is, that the closed unit Ball $\bar{B}_{1}^{\prime\prime}\subseteq E^{\prime\prime}$ is w*-separable and w*-compact, and thus w*-Polish.
Question: is $E^{\prime\prime}$ indeed w*-separable? And is my corollary valid?
You are correct that $E''$ is weak*-separable. However, your corollary does not follow: a separable compact space need not be metrizable, and thus need not be Polish. Indeed, your corollary is false. For instance, if $E=\ell^1$, then $E''=(\ell^\infty)'$ can be identified with the space of regular Borel complex measures $M(\beta\mathbb{N})$ on the Stone-Cech compactification $\beta\mathbb{N}$ of $\mathbb{N}$. Since $\beta\mathbb{N}$ has cardinality $2^{2^{\aleph_0}}$, so does $M(\beta\mathbb{N})$ (consider just the point measures), and thus so does the unit ball in $M(\beta\mathbb{N})$. A separable metric space has cardinality at most $2^{\aleph_0}$ (since every point is the limit of some sequence in the countable dense subset), so the unit ball cannot be weak*-metrizable.