$\newcommand\vpi\varphi\newcommand\R{\mathbb R}$
The approach involving (4) will not work, because Lévy's continuity theorem will guarantee that the pointwise limit of characteristic functions (c.f.'s) is a c.f. only when the limit function function is continuous (everywhere or, equivalently, at the origin).
Nonetheless, your c.f. $\vpi_\mu$ is nonzero everywhere. Indeed, for any $x'\in E'$ and any real $t$, $$\vpi_\mu(tx')=\int_E\mu(dx)\,e^{itx'(x)}=\vpi_{x'\sharp\mu}(t),\tag{1}$$ where $\vpi_{x'\sharp\mu}$ is the c.f of the probability measure $x'\sharp\mu$ over $\R$ that is the pushforward of $\mu$ under the map $x'$. The probability measure $x'\sharp\mu$ over $\R$ is infinitely divisible, since $\mu$ is infinitely divisible: if $\mu=\mu_n^{*n}$, then $x'\sharp\mu=(x'\sharp\mu_n)^{*n}$. So, by the Lévy--Khintchine formula, $\vpi_{x'\sharp\mu}(t)\ne0$ for all real $t$. Taking now $t=1$ in (1), we get $\vpi_\mu(x')\ne0$ for all $x'\in E'$, as desired.