Natural transformations induce homotopies - Is this true in the “fat” world?

Question

Let $\mathcal{C}, \mathcal{D}$ be categories internal to topological spaces (or compactly generated Hausdorff spaces, if you like) $F,G\colon\mathcal{C}\rightarrow\mathcal{D}$ be continuous functors and $\eta\colon F\implies G$ a natural transformation, i.e. a continuous functor $$\mathcal{C}\times\{0,1\}\rightarrow \mathcal{D}.$$

Using the ordinary geometric realization, this functors induces a homotopy between $BF$ and $BG$ since $B$ preserves products and $B(\{0,1\})\cong[0,1]$.

If one uses the fat realization $B'(\_)$, products are probably preserved up to homotopy, but $B'(\{0,1\})$ will be something infinite dimensional. Is it still true that $B'F$ and $B'G$ are homotopic?

If the nerves of $\mathcal{C}$ and $\mathcal{D}$ are "good" in the sense of Segal, the ordinary realization coincides up to homotopy with the usual one and the question can be answered positively. However, is it still true in the "bad" world?

Edit: The fact that products are preserved up to homotopy is Proposition A.(iii) of Segal's "Categories and Cohomology Theories".

Answer 1

As pointed out by the OP, fat realisation preserves products up to homotopy, so we get a map $$B'(C)\times B'(\{0\lt1\}) \simeq B'(C\times \{0\lt1\}) \to B'(D).$$ The space $ B'(\{0\lt1\})$ is bipointed and path-connected (in fact contractible) so we get a homotopy $B'(C)\times [0,1]\to D$. Then one needs to show the maps it is a homotopy between are themselves homotopic to the two induced maps $B'(C\times \{i\})\to B'(D)$, $i=0,1$.