Is there a convergent, alternating series that fails the AST?

Yes, there are such series. Consider, for example, a sequence such as

$$1, 2, \frac 1 2, 1, \frac 1 4, \frac 1 2, \frac 1 8, \frac 1 4, \dots$$

The series

$$1 - 2 + \frac 1 2 - 1 + \frac 1 4 - \frac 1 2 + \frac 1 8 - \frac 1 4 + \dots$$

is alternating and (absolutely) convergent, but it clearly fails to be monotonically decreasing.

Let $a_n = \begin{cases}2^{-n} & n \equiv 0\pmod{2} \\ 3^{-n} & n \equiv 1\pmod{2} \end{cases}$. Then, $a_n$ is not strictly decreasing, since $a_3 = \dfrac{1}{27} < \dfrac{1}{16} = a_4$.

However $\displaystyle\sum_{n = 0}^{\infty}(-1)^na_n$ still converges to $\dfrac{1}{1-\tfrac{1}{4}} - \dfrac{\tfrac{1}{3}}{1-\tfrac{1}{9}} = \dfrac{23}{24}$.

Pick any series $\sum (-1)^na_n$ that satisfies the AST, and pick any sequence $b_n$ that converges to $0$.

Define

$$c_{n}=a_n+b_{\lfloor \frac{n}{2} \rfloor}$$ that is $$c_{2n}=a_{2n}+b_n \\ c_{2n+1}=a_{2n+1}+b_n$$

Then, it is easy to prove that $\sum(-1)^n c_n$ is always convergent, but it is very easy to make examples where $c_n$ is not positive, or not decreasing. Or to fail both conditions.