The claim fails in $\textbf{Top}^\textrm{op}$, amusingly. Let $A = \{ a, b, c, d \}$ with open sets $\emptyset, \{ a \}, \{ b \}, \{ a, b \}, \{ a, c \}, \{ b, d \}, \{ a, b, c \}, \{ a, b ,d \}, \{ a, b, c, d \}$ let $B = \{ 0, 1, 2 \}$ with open sets $\emptyset, \{ 0 \}, \{ 0, 1 \}, \{ 0, 1, 2 \}$ and let $C = \{ 0, 2 \}$ be topologised as a subspace of $B$. Let $p : A \to B$ be the continuous map given by $p(a) = 0, p(b) = 1, p(c) = 1, p(d) = 2$ and let $i : C \to B$ be the inclusion. It is not hard to see that $p$ is a quotient map, so it is a regular epimorphism; now consider the pullback of $p$ along $i$. This is the map $q : D \to C$ where $D = \{ a, d \}$ is a discrete subspace of $A$, and $q$ is certainly surjective (so is an epimorphism) but $q$ is not a quotient map (so not a regular epimorphism). Thus, pullbacks in $\textbf{Top}$ do not preserve regular epimorphisms; equivalently, pushouts in $\textbf{Top}^\textrm{op}$ do not preserve regular monomorphisms.