On page 227 in Borceux's Categorical Algebra I, he defines a universal closure operation on a finitely complete category $\mathscr B$ to be an operation on $\overline \square\colon \mathbf{Sub}(B)\to \mathbf{Sub}(B)$ for each $B\in \mathscr B$ that is extensive $S\subseteq \overline S$, monotone $S\subseteq T\Rightarrow \overline S\subseteq \overline T$, idempotent $\overline{\overline S}=\overline S$, and closed under inverse image $\forall f\colon A\to B$ in $\mathscr B$, $f^{-1}(\overline S)=\overline{f^{-1}(S)}$.
He immediately gives the following proposition:
If $\mathscr B$ is finitely complete with a universal closure operation, then for all sub objects $S,T$ of $B\in \mathscr B$, $\overline{S\cap T}=\overline S \cap \overline T$.
If we let $\mathbf{Top}=\mathscr B$, $\mathbb R=B$, $(0,1)=S$, $(1,2)=T$, with the usual closure operator, then $\overline S \cap \overline T\neq \overline{S\cap T}$. How is this not a counterexample? Am I missing something from the definition of sub objects?
The usual closure operator on subsets of topological spaces doesn't satisfy the last condition: in general, $f^{-1}(\overline{S})$ can be strictly larger than $\overline{f^{-1}(S)}$. For example, $f$ might have the property that its image doesn't intersect $S$ but does intersect the closure of $S$.