A homotopy between proper maps is automatically proper

This is smooth. At both endpoints, $H(x,0)=x=H(x,1)$, so $H_0$ and $H_1$ are the identity of $\mathbb{R}$. Each is proper because the inverse image of every compact set is that same compact set.

The singleton $\{0\}\subset \mathbb{R}$ is compact, but $H^{-1}(\{0\})=(\{0\}\times [0,1])\cup (\mathbb{R}\times \{1\mathrm{/}2\})\mathrm{.}$ Its closed subspace $\mathbb{R}\times \{1\mathrm{/}2\}$ is homeomorphic to the noncompact real line from [F2]; more directly, the cover by $(-n,n)\times (1\mathrm{/}4,3\mathrm{/}4)$ together with the complement of that slice has no finite subcover. Hence $H^{-1}(\{0\})$ is not compact, and $H$ is not proper.

Thus proper endpoint maps do not make the combined homotopy proper, and [F1]'s hypothesis cannot be deleted. The midpoint map is the constant zero map, which pinpoints the degeneration. Both endpoints, the compact singleton, and the noncompact inverse image are explicit; the source is nonempty and boundaryless in its spatial variable, and no choice principle is used.