Let's look at the observed facts on radioactivity first.
Radioactive decay is not observed to be a process in which (on average) a set proportion of the original mass at some initial time $t = 0$ is lost over equal time periods until it loses all radioactivity.
By continually measuring the radiation intensity from a given mass of radioactive material with some suitable Geiger counter, you won't observe a linear drop-off to an eventual zero (or ambient background intensity level) radiation. You will see a fall-off in intensity whereby the loss in radiation intensity per unit time is itself continually decreasing with time from the start of measurement.
This observed curve instead signifies a proportional loss mechanism whereby a proportion of the current radioactive mass is being lost continually throughout its full radioactive life.
So:
$$ -\frac{dN}{dt} \propto N $$
Rearranging:
$$ \frac{dN}{N} = - \lambda dt $$
where $\lambda$ is a constant signifying the proportional rate of radioactive decay with dimensions like percent per second.
This can be integrated and leads us to:
$$ N(t) = N_0 e^{-\lambda t} $$
where $N_0$ is the original number of radioactive atoms at some initial time. This function has the shape of the graph above.
Regardless of what size $N_0$ may be, when the left-hand side of this equation is some known fraction of $N_0$ then the value of $t$ depends only on the decay-proportion rate ($\lambda$) and that fraction of $N_0$ that we have on the LHS - both of which are constant for a given substance.
For example, if $N(t) = N_0 / 2 $ we can get the time for half the radioactive mass present at $t = 0$ to decay (i.e. the half-life) via:
$$ t_\frac{1}{2} = \frac{ln2}{\lambda} $$
The assertion that the second half of a radioactive substance's mass should need no longer than its half-life to fully expire its radioactivity is clearly not applicable here. As explained, the second half of the radioactive mass will be lost at an ever reducing rate and thus take much longer. In fact, to be exact about it, the remaining mass will take forever to lose full radioactivity as its decay rate slows down to near nothing towards the end.
The situation is analogous to unfilling a unit square by continually taking half of the square's remaining area. In this way after taking $n$ withdrawals we have a remaining area of:
$$ A = 1 - (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{2^n}) $$
As $n$ becomes large, the removed area tends to unity and the remaining area to zero. But this process never fully removes all the remaining area - it just takes half of whatever is left.
So it takes an infinite number of withdrawals to completely unfill the unit square - although of course we can reduce it to a very small size by a relatively small number of extractions.
The notion of half-life to measure the persistence of a radioactive source is useful insofar as it gives us a relatable means of describing how long a radioactive substance is significantly present and emitting potentially dangerous radiation to the atmosphere. If we are happy with 0.1 % of the radioactive mass being there and know its half-life to be 3 months, we can find the number of half-lives $N$ required to achieve this reduction via:
$$ (\frac{1}{2})^n = \frac{0.125}{100} $$
$$ \Rightarrow n \approx 10 $$
So after 10 x 3 months $\approx 2.5$ years there will be less than 0.1 % of the radioactive mass still there.