I'm more familiar with the saying as "a stopped clock is right twice a day", which is obviously correct for a standard 12 hour analog clock (with a possible exception involving leap-seconds).

What I would consider to be the key insight is that, if you have two clocks running at (different) constant speeds, τ, ω, then the difference between them also changes at a constant rate, |τ−ω|. It follows immediately that, if the first clock is correct and working properly, then τ=1, and the difference of the other from the correct time, t, is given by |1−ω|t.

Since clocks are periodic, if the broken clock is correct at t=0, it will also show the correct time at any t such that |1−ω|t=(12hours)×k, or t=(12hours)×k|1−ω| (for k∈Z). The period of that, T, is clearly given by T=12hours|1−ω|, and, since ω can be made as close as you choose to 1, T can be made as close to ∞ as you wish.

That also makes intuitive sense. If a clock loses or gains time very slowly - say a second per year - then it also takes a very long time for that error to be noticeable, and a much, much longer time before that error accumulates to be 12 hours.

Meanwhile, a clock that's running at a high enough speed (forward or backward) will display every time multiple times per second, including the correct time, so, at least in theory, for any short, finite time interval, δt, you can describe a clock that runs continuously at a fixed speed, is correct at some instant during the interval [t,δt) for any time t, and which is completely useless in practice.