 Solution

Put $L={\bigcap }_{i=1}^k{H}_i$. Repeated application of part (iv), or mathematical induction on $k$, givesPart (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: