33.39 Let $x, \quad y \in R, \quad|x|<1, \quad|y|<1, \quad x y \neq 0$. Denote $[x]$ as the greatest integer not exceeding the real number $x$. Then the solution set region for the inequality $[x+y] \leqslant[x]+[y]$ (as shown in the figure) is
[Solution] Take $x=\frac{1}{3}, y=\frac{2}{3}$, then $[x+y]=1$, and $[x]=[y]=0$, at this time $[x+y]>[x]+[y]$, which contradicts $[x+y] \leqslant[x]+[y]$, but $\left(\frac{1}{3}, \frac{2}{3}\right)$ is in the interval shown in figures $(B),(D)$, so $(B),(D)$ are not suitable.

Take $x=-\frac{1}{2}, y=\frac{1}{4}$, then $[x+y]=-1$, and $[x]=-1,[y]=0$, at this time $[x+y]=[x]+[y]$, which satisfies the given inequality.

But $\left(-\frac{1}{2}, \frac{1}{4}\right)$ is not in the interval shown in figure $(A)$, so $(A)$ is not suitable.
Therefore, the answer is $(C)$.