group of units in a topological ring

(I've never offered a bounty before; I hope it's not bad etiquette to post a solution now, even though technically we're in a "grace period". Since there has been no activity, I'm guessing it'll be alright.)

There is a counterexample, suggested in Warner's book (linked in the question), p. 113, involving a ring topology on the rational numbers $\mathbb{Q}$. Take basic neighborhoods of the origin to be ideals of $\mathbb{Z}$, i.e., subsets of $\mathbb{Q}$ of the form $n\mathbb{Z}$ where $n \neq 0$ is an integer. By taking lcm's, the collection of such sets is closed under finite intersections. Thus a basis of the topology consists of sets of the form $q + n\mathbb{Z}$ where $q \in \mathbb{Q}$. That this gives a ring topology (i.e., a topology for which subtraction and multiplication on $\mathbb{Q}$ are continuous) may be boiled down to three easily checked claims (cf. Warner, theorems 11.2, 11.3, and 11.4, pp. 78-79):

• For each ideal $I$ there is an ideal $J$ such that $J + J \subseteq I$ and $-J = J$ (obviously $J = I$ will do);

• For each ideal $I$ and $r \in \mathbb{Q}$, there is an ideal $J$ such that $rJ \subseteq I$ (if $r = p/q$ for integers $p, q$, then $J = qI$ will do);

• For each ideal $I$ there is an ideal $J$ such that $J \cdot J \subseteq I$ (again $J = I$ will do).

Also notice that $\{0\}$ is closed in this topology, because if $q \neq 0$ and we pick $n \in \mathbb{Z}$ with $|n| > |q|$, then the open neighborhood $q + n\mathbb{Z}$ of $q$ doesn't contain $0$.

Note reciprocation = multiplicative inversion is not continuous w.r.t. this topology. Indeed, for the neighborhood $1 + 2\mathbb{Z}$ of $1$, its inverse image under reciprocation is the set $\{\frac1{1 + 2n}: n \in \mathbb{Z}\}$, and no basic open neighborhood $1 + m\mathbb{Z}$ of $1$ is small enough to fit inside this set. Thus, we have exhibited a counterexample.