The definition of proper homotopy $H: [0,1]\times X\rightarrow X$ requires that $H(t,x)$ is proper as a full map (from product $[0,1]\times X$ to $X$).
I feel that it is not equivalent to that $H(t,\cdot): X\rightarrow X$ is proper for each fixed $t\in [0,1]$ (on the other hand, if $X$ is compact, both cases are true).
Do you know any counterexample, when for each fixed $t$, $H_t(x) = H(t,x)$ is proper but the full map is not?
Let $X=\mathbb R.$ The homotopy will be from the identity map to itself, so $H(0,x)=H(1,x)=x$ for all $x.$ For each integer $n\geq 1,$ during the time period $[1/(n+1),1/n],$ the point $n\in X$ is taken to $0$ and back, and points outside $[n-1,n+1]$ are not affected ($H(t,x)=x$ for $x\not\in[n-1,n+1]$ and $t\in[1/(n+1),1/n]$). Then $H^{-1}(\{0\})$ is unbounded, but $H_t^{-1}(\{0\})$ is bounded for each $t.$