The terminology is unfortunately misleading. The event of interest is not infection per se, but transmission of the virus. The distinction in language is subtle but important. An individual is considered infected if the virus is present, but a disease may be transmitted on more than one occasion. For a disease with effectively no cure, infection may be regarded as a permanent state--i.e., once infected, one remains so--but transmission is a person-to-person event that may occur any number of times. Therefore, the event that the author speaks of is not the former, but the latter, and $A_1$, $A_2$ should more precisely be regarded as transmission events.
With this clarification, it becomes obvious that if each sexual encounter carries an independent and identically distributed probability $p$ of resulting in transmission of the virus, then $$\Pr[A_1 \cup A_2] = \Pr[A_1] + \Pr[A_2] - \Pr[A_1 \cap A_2] = 2p - \Pr[A_1]\Pr[A_2] = 2p - p^2 < 2p.$$ If $\Pr[A_i] = p$ for each encounter $i$--that is to say, identically distributed but not necessarily independent--then $\Pr[A_1 \cap A_2]$ is not known, but as long as they are not mutually exclusive, the joint probability is strictly positive, thus the inequality stated by Rice holds. And this is where you had difficulty following due to his unfortunate use of terminology, because you correctly infer that if $A_i$ denote infection events, then this probability becomes zero, since $A_1 = 1$ implies $A_2 = 0$.
The natural extension of the iid model assumptions tells us that the outcome of each encounter is an independent and identically distributed Bernoulli random variable, say $$A_i \sim \operatorname{Bernoulli}(p), \quad \Pr[A_i = 1] = p, \quad \Pr[A_i = 0] = 1-p$$ where $A_i = 1$ if the outcome is transmission of the virus, and $A_i = 0$ if the outcome is non-transmission. For an individual with $n$ such encounters, we let $X$ be the number of transmission events experienced, hence $$X \sim \operatorname{Binomial}(n, p), \quad \Pr[X = x] = \binom{n}{x} p^x (1-p)^{n-x}, \quad 0, 1, \ldots, n,$$ and this individual is considered infected if $X \ge 1$. Consequently the probability that an individual remains uninfected after $n$ encounters is $$\Pr[X = 0] = (1-p)^n,$$ and the complementary probability--i.e., the individual becomes infected within $n$ encounters, is $$\Pr[X \ge 1] = 1 - \Pr[X = 0] = 1 - (1-p)^n.$$
The original article also makes some sweeping assumptions about transmission probabilities as well as an unwarranted assumption about the independence and person-to-person homogeneity of transmission events, but since we lack source data to model these in a more sophisticated manner, the best we can do is to note that the average of the claimed "number needed to transmit"--what the article calls "risk"--is inappropriately computed; it is not the simple arithmetic mean of $1/100$ and $1/1000$ as claimed. Rather, a geometric mean of these rates is a better estimate of the average rate of transmissions per encounter; i.e., $$p = \sqrt{10^{-2} \cdot 10^{-3}} \approx 0.00316228 \approx \frac{1}{316}.$$ The reason for this is left as an exercise for the reader.
As to the last part of the article quoted, "...100% probability of infection..." this is patently absurd. It is equivalent to saying that you are guaranteed to get at least one head if you flip a fair coin twice, because there are only two distinct outcomes for a coin toss. This illustrates the true nature of the extent of innumeracy of the article's author.
