I need to find supremum of the set of all real numbers of the form $\operatorname{Re}(iz^3+1)$ such that $|z|<2$.
By the inequality $-|w|\le \operatorname{Re}(w)\le |w|$ we have $$\operatorname{Re}(iz^3+1)\le |iz^3+1|<|i||z^3|+1<9$$ but then can I say the sup is $9$?
So far so good: you correctly found that $9$ is an upper bound.
It's probably not going to be attained since the inequality $|z|<2$ is strict. Imagine for a moment we had $|z|\le 2$. Then $iz^3$ would be equal to $8$ when $z=-2i$, thus delivering $iz^3+1=9$.
Back to reality: $z=-2i$ is not allowed. But, you can take $z=-iy$ for any $0<y<2$. As $y$ approach $2$, $iy^3+1$ approaches $9$. The fact that supremum is exactly $9$ follows from this, but you may need to polish this a bit depending on the level of detail expected from you.