(I am most interested in the case $X=\mathbb R^2$, but of course one could ask the same question for manifolds, or metric spaces in general.)


Let $\text{Com}(\mathbb R^2)$ denote the space of nonempty compact subsets of the plane, equipped with the Hausdorff metric. Let $S\_\bullet:[0,1]\to\text{Com}(\mathbb R^2)$ be a continuous path, and let $p\in S\_0$. Must there exist a path $\gamma:[0,1]\to\mathbb R^2$ such that $\gamma(0)=p$, and $\gamma(t)\in S\_t$ for all $t\in[0,1]$?



Must a path of compact sets in $X$ descend to a path in $X$?
$\DeclareMathOperator{\R}{\mathbf{R}}\DeclareMathOperator{\Z}{\mathbf{Z}}$The answer is no, even in the circle (and hence in the plane).


As coordinates, write the circle as the 1-point compactification $\bar{\R}$ of $\R$.


For $t\in\mathopen]0,1]$, write $$X\_t=\{\infty\}\cup\big(t\Z+\sin(1/t)\big).$$
For $t\to 0$, this tends to $X\_0=\bar{\R}$, hence defines a continuous path on $[0,1]$. Any continuous lift, for $t>0$ has to have the form $x(t)=tn+\sin(1/t)$ for some fixed $n$. This does not converge when $t\to 0$ (it accumulates to all of $[-1,1]$). So there is no continuous lift.


(Note that the answer is clearly positive when $X=\mathbf{R}$, as $x\mapsto \max(x)$ is then a lift (not only for paths).)


