The Intervall should $(-\infty,0]$, Since the right limit of log at zero is $-\infty$
Edited due to dsaxton's comment:
Note that the probability that U takes any value $c \in [0,1]$ is 0, or in math notation $P(U=c)=0$ for every $c \in [0,1]$
Why do we take limits, (do we?)?
First, let me remark that log is bijective strictly increasing function from $(0,1] \to (-\infty,0]$. Using some math one could define the continuous extension of in zero log to be $log(0):= -\infty$
However, in that case, $P(X=-\infty)=P(U=0)=0$, and the value $-\infty$ won't play a role.
However, to determine the image of $X$, we employ the range of log on $(0,1]$ being $(-\infty,0]$. In both intervals, and in the determination of the image of log on $(0,1]$ limits are involved