$\sup_{n} E[|X_{n}|\mathbb{1}_{[X_{n}|>c}]\rightarrow0$ as $c \rightarrow \infty$, being the definition of uniform integrability, should not occur if the martingale converges? Why is that?
Further, one apparently can argue that $E[X_{n}]=c \not=0 \Rightarrow$ non-uniformly integrable, though I do not see also how this should work (the indicator function within the expectation I thought would have allowed $E[X_{n}]=c$ to not negate u.i.)
COMMENT 1 : See, a uniformly integrable martingale which converges almost surely, converges in $L^1$ as well. As a consequence, whatever random variable $X_n$ converges a.s. to, must have the same expectation as $X_0$. Now, if it is clear from the construction of the $X_n$ that you are looking at, that the limit does not have the same expectation as $X_0$ (or any of the $X_i$) then "clearly" the martingale is not UI, where "clearly" is to the author(and to you now)!
COMMENT 2 : The definition of UI is not used in the "clearly" part, it seems. Instead the following fact is used : given a sequence of (integrable) random variables $X_n$ and a random variable $Y$, the fact that $X_n \to Y$ in probability and $X_n$ is UI, is equivalent to $X_n \to Y$ in $L^1$. In your context, I want you to find the $Y$ which the $X_n$ goes to a.s. and hence in probability. Maybe by looking at the expectations you can see that $X_n \to Y$ in $L^1$ is not possible, and therefore $X_n$ cannot be UI.