I just happened to notice that "counital" is currently not settled for the category of Banach spaces with linear contractions. After a bit of thought, I came up with this possible proof, which I'm not confident in though:
Suppose we have an epimorphism $A \twoheadrightarrow B$ with maps $f : A \to X \sqcup Y$ and $g : B \to X \times Y$ making a commutative square. Then the function $B \to \mathbb{R}$, $b \mapsto |b|_B - |p_1(b)|_X - |p_2(b)|_Y$, is a continuous function, and it's non-negative on the image of $A$. Since the image of $A$ is dense in $B$, that implies the function is non-negative everywhere on $B$, implying that $g$ factors through $X \sqcup Y$. That establishes that $X \sqcup Y \to X \times Y$ is a strong monomorphism.
And assuming that's correct, then that would give lots of examples of strong monomorphisms which are not regular, implying that Ban is not coregular.
I just happened to notice that "counital" is currently not settled for the category of Banach spaces with linear contractions. After a bit of thought, I came up with this possible proof, which I'm not confident in though:$A \twoheadrightarrow B$ with maps $f : A \to X \sqcup Y$ and $g : B \to X \times Y$ making a commutative square. Then the function $B \to \mathbb{R}$ , $b \mapsto |b|_B - |p_1(b)|_X - |p_2(b)|_Y$ , is a continuous function, and it's non-negative on the image of $A$ . Since the image of $A$ is dense in $B$ , that implies the function is non-negative everywhere on $B$ , implying that $g$ factors through $X \sqcup Y$ . That establishes that $X \sqcup Y \to X \times Y$ is a strong monomorphism.
Suppose we have an epimorphism
And assuming that's correct, then that would give lots of examples of strong monomorphisms which are not regular, implying that Ban is not coregular.