This might be a stupid question, but since a topology on a set $X$ is a collection of subsets of $X$, $\tau$, where for each element, $U \in \tau$, $U$ is an open set, and since there is also the requirement $X\in \tau$, how does this work where $X$ is a closed set like $[0, 1]$?
Also, since the complement of an open set is a closed set, does $X^c \neq \emptyset$?
You are wrong about the definition of topological space. Such a space is a pair $(X,\tau)$, where $X$ is a set and $\tau$ is a subset of $\mathcal{P}(X)$, for which certain conditions must hold ($\emptyset,X\in\tau$ and $X$ is stable with respect to arbitrary unions and to finite intersections). Only after that we define open sets. It's just this: the open sets of a topological space $(X,\tau)$ are the elements of $\tau$. Besides, the closed sets are those $Y\subset C$ such that $Y^\complement\in\tau$.
So, yes, we can perfectly define topologies $\tau$ on $\mathbb R$ for which $[0,1]$ is an open set. And for such a space, $[0,1]$ may be or not a closed set.
By the way, keep in mind that “open” is not the opposite of “closed”: a subset of a topological space may be open and closed at the same time (the whole space and the empty set, for instance) and may also be neither open nor closed.