Sampling at subnyqvist frequencies
The point in the article is that you can sample the signal at lower frequencies than 60Hz by subsampling. Consider the following two scenarios of sampling the 60Hz signal:


You sample like Nyquist wanted, at freq $\geq 120\mbox{Hz}$. In one period of your 60Hz signal, or $T=16.67\mbox{ms}$, you will have a correct representation of your signal via sampling.





You sample at a frequency even lower than the $60\mbox{Hz}$ signal freq. The trick is to sample at frequencies that do not divide the main frequency, which means you cannot use $2, 3, 6, 10, 12, 15, 20, \mbox{ or } 30 \mbox{Hz}$ frequencies, because then you'd always get the same point of the signal.


Let's say you sample at $19\mbox{Hz}$ like in the article. The whole point of this is to save the points on your graph, and get the desired signal point-by-point. So you start sampling your signal, get one point, then whole 3 periods of the signal will pass, and you get a second point, which you put on the graph, and so on until you have enough. 

How many points do we need? The Nyquist theorem you stated tells us the number of points we need to be sure we got 100% of the signal. $$T_s = (M + \frac{P}{N})\frac{1}{F},$$  where $F$ is our signal frequency of $60\mbox{Hz}$, $T_s = 1/F_s = 1/19\mbox{s} = 52.63\mbox{ms}$ is our sampling period, and $M, N, P$ are integers where $N$ and $P$ must have no common factors. If we try to get such integers, we can see that for $N = 19, M = 3, \mbox{ and } P = 3$ our equation holds. So, we need 19 points for our graph, in order to have the whole signal. This means we need to wait 19 times our sampling period to get our total time of sampling, so we get $$ T_{tot} = 19\cdot 52.63\mbox{ms} = 999.97\mbox{ms} $$



So we see that sampling is possible at even lower frequencies, but at much longer time :) Hope this clarified your question! :)