= Solution
{c}
Put $L=\bigcap_{i=1}^kH_i$. Repeated application of part (iv), or <mathematical induction> on $k$, gives
$$
|G/L|\le\prod_{i=1}^k n_i<\infty.
$$
Part (ii) therefore implies that $L$ is an <infinite set>. If $G\setminus U$ is nonempty, choose $g\notin U$. We claim its <left coset> $gL$ is wholly outside $U$. If $g\ell\in H_i$ for some $\ell\in L$, then $\ell\in H_i$ and closure under products and inverses would give $g=(g\ell)\ell^{-1}\in H_i$, a contradiction.
Thus $gL\subseteq G\setminus U$. Part (i) makes $gL$ infinite, so the complement is infinite. If there is no such $g$, the complement is empty. This proves the <complement of a finite union of finite-index subgroups> property:
$$
\boxed{G\setminus(H_1\cup\cdots\cup H_k)\text{ is empty or infinite}.}
$$
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