double comb space is not contractible

Have you tried using uniform continuity? $D \times I$ is a compact metric space so the function $H$, if it is continuous, must also be uniformly continuous. This could have implications for how $H$ maps two sets $x \times I$ and $y \times I$ when $x,y \in D$ are very close to $x_0$ but $x$ is to the left and $y$ is to the right of $x_0$.

Let $x_0=(0,1)$, the point where the two combs are wedged.

There is $t_0$ such that

$H(x_0\times[0,t_0])=\{x_0\}$ and for each $\epsilon>0$ there is an $0<\alpha<\epsilon$ and a $\lambda\ne0$ with $H(x_0,\ t_0+\alpha)=(0,\ 1+\lambda)$

Take $0<\epsilon<1$. By continuity, there is a $\delta>0$ with $$H(B_δ(x_0)×[0,\ t_0+δ])\subseteq B_ϵ(x_0)$$ There is an $0<\alpha<δ$ with $$H(x_0,\ t_0+α)=(0,\ 1+\lambda)$$ where $λ\in(-ϵ,ϵ)$. Again by continuity, there is a $0<\beta<α$ such that $$H(B_β(x_0)×[-β+(t_0+α),\ (t_0+α)+β])\subseteq B_λ(0,\ 1+λ)$$ Assume without loss of generality that $λ>0$. But what does this imply for a point $y=(y_1,1)$ where $0<y_1<β$. It means that $y$ must travel along a path to a point in the upper comb (since $B_λ(0,1+λ)$ doesn't meet the lower comb), where it arrives at some time $t\in((t_0+α)-β,\ (t_0+α)+β)$. To do so, it had to go down to $I×\{0\}$, which is impossible since $H(B_δ(x_0)×[0,\ t_0+δ])⊆B_ϵ(x_0)$.

There is still the possibility that $t_0=1$ and the point $(0,1)$ is fixed during the $I$. But that would mean that the comb space strongly deformation retracted onto this point. But then for some small ball $B$ around $x_0$, some smaller ball had to be stay in $B$ during the entire time, preventing the point $y_1$ from travelling along the path through the base.