From Williams' Probability with Martingales
How do we know that $A_{\infty} < \infty$?
If $T = \infty$, then
$$E[A_{T \wedge n}] \le (K+c)^2$$
$$\to E[A_{n}] \le (K+c)^2$$
$$\to \lim E[A_{n}] \le (K+c)^2$$
$$\to E[\lim A_{n}] \le (K+c)^2$$
$$\to E[\lim A_{n}] < \infty$$
$$\to \lim A_{n} < \infty$$
$$\to A_{\infty} < \infty$$
If $T < \infty$, then
$$E[A_{T \wedge n}] \le (K+c)^2$$
$$\to \lim E[A_{T \wedge n}] \le (K+c)^2$$
$$\stackrel{Why?}{\to} E[\lim A_{T \wedge n}] \le (K+c)^2$$
$$\to E[A_{T}] \le (K+c)^2$$
I'm stuck. How can I approach this?
You've shown that there is a $c$ such that $P(T=\infty)>0$. So:
$$E[A_{T\wedge n}]=E[A_{T\wedge n}|T<\infty]P(T<\infty)+E[A_{T\wedge n}|T=\infty] P(T=\infty)$$
$$\to E[A_{T\wedge n}]=E[A_{T\wedge n}|T<\infty]P(T<\infty)+\frac{E[A_{T\wedge n}1_{T=\infty}]}{P(T=\infty)} P(T=\infty)$$
$$\to E[A_{T\wedge n}]=E[A_{T\wedge n}|T<\infty]P(T<\infty)+\frac{E[A_{n}1_{T=\infty}]}{P(T=\infty)} P(T=\infty)$$
$$\to E[A_{T\wedge n}]=E[A_{T\wedge n}|T<\infty]P(T<\infty)+\frac{A_{n}E[1_{T=\infty}]}{P(T=\infty)} P(T=\infty)$$
$$\to E[A_{T\wedge n}]=E[A_{T\wedge n}|T<\infty]P(T<\infty)+A_n P(T=\infty)$$
Both terms terms on the r.h.s are non-negative and you've also shown that $E[A_{T\wedge n}]$ is bounded above and you can take $n\rightarrow\infty$, so at the very least $A_\infty<\infty$ since $P(T=\infty)>0$