= Solution
Under the temporary assumption that $E$ is a <basis of a module> for $B$, part (a)(v) applied to the countably generated <submodule> $F$ gives a <countable set> $E'\subseteq E$ such that $F\subseteq C=\langle E'\rangle$. By part (a)(iv),
$$
B=C\oplus\langle E\setminus E'\rangle.
$$
Use part (a)(iii) to project onto the second <direct summand>. This gives a <module homomorphism> $\pi:B\to B$ with
$$
\boxed{F\subseteq\ker\pi=C}.
$$
It remains to check the required size of this <kernel>. Enumerate $E'$ as $b_1,b_2,\ldots$, with the finite case treated the same way. Every element of $C$ is a finite sum of these elements with <integer> coefficients. For each $m$, the set $\langle b_1,\ldots,b_m\rangle$ is the image of the <countable set> $\mathbb Z^m$, so it is <countable>. Taking their <countable union> proves
$$
\boxed{\ker\pi\text{ is countable}}.
$$
This is the <countability of countably generated abelian groups> argument; it uses the countability of the scalar <ring> $\mathbb Z$.
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