For continuous random variables: $E(U'(\xi)) = \int_0^{+\infty} U'(x) f_\xi(x)dx < +\infty$, because $U'(+\infty)=0$, and boundness of $U'$ is suffucuent here.

Then, $\xi U'(\xi) - U'(\xi) \leq U(\xi) - U(1)$, because $U$ is concave, and $E(\xi U'(\xi)) \leq E(U(\xi)) + E(U'(\xi)) - U(1) < +\infty$.

Take $U(x) = -e^{1/x^2}$ and the probability density function $f(x) = C e^{-1/x^2}$, $x \in [0;1]$, where $C$ is normalization constant.

Then, $U'(x) = \frac{2 e^{1/x^2}}{x^3}$, $E(U(\xi)) = \int_0^1 U(x) f(x) dx = -C$, $E(\xi U'(\xi)) = \int_0^1 x U'(x) f(x) dx = 2C \int_0^1 \frac{dx}{x^2} = +\infty$