Does every compact simply-connected subset of $\mathbb{R}^n$ have an efficient $r$-covering path for all $r>0$? Let $A$ denote a subset of $\mathbb{R}^n$.
Definition 0. Given a positive real number $r$, an $r$-covering path of $A$ is a non-negative real number $T$ together with a differentiable function $c:[0,T] \rightarrow \mathbb{R}^n$ satisfying: for all $a \in A$, there exists $t \in [0,T]$ such that $\|a - c(t)\| \leq r.$
Remark. I'm not especially attached to the details of the above definition. For example, you may wish to replace "differentiable" with "rectifiable." Or, you may wish to include an assumption that $c'(t) \neq 0$ for all $t \in [0,T],$ in order to remove the possibility of "kinks" from the path. So basically, use whatever definition makes the question easier to answer in the affirmative.
Definition 1. An $r$-covering path of $A$ is efficient iff there is no $r$-covering path of strictly shorter length.
I reckon the following question is pretty interesting, especially if the answer is "yes."

Question. Is it true that for all $r>0$, every compact simply-connected subset of $\mathbb{R}^n$ has an efficient $r$-covering path?

The conditions "compact and simply-connected" were basically plucked out of thin air to in order to eliminate weird and silly counterexamples. Anyway, feel free to modify them as you see fit. The same goes for "differentiable": feel free to replace this with "smooth" or even "analytic" if it makes for a cleaner answer.
 A: Rectifiable case
The comment by Nate Eldredge solves this one. Indeed, let $L$ be the infimal length of $r$-covering paths, and pick a sequence of such paths $\gamma_n$ of length at most $ L+1/n$. Each of them can be represented by an $(L+1/n)$-Lipschitz map from $[0,1]$. These maps are also uniformly bounded, since the paths visit the $r$-neighborhood of the same bounded set. By the Arzelà-Ascoli theorem, there is a uniformly convergent subsequence $\gamma_{n_k}$. Its limit $\gamma$ has length at most $   L$, since length is lower semicontinuous under uniform convergence. It also visits the $r$-neighborhood of every point of $A$. Hence, the length of $\gamma$ is $L$, and it is an efficient $r$-covering path.
Note that we don't need any assumptions on $A$ other than boundedness.
Smooth case
Within differentiable curve, the infimal length need not be attained. For example, let $A\subset \mathbb R^2$ be the broken line from 
$A(0,3)$ to $B(0,0)$ to $C(3,0)$, and let $r=\sqrt{2}$. There is a non-differentiable $r$-covering path of length $2(\sqrt{5}-\sqrt{2})$, shown below: it is obtained
from the broken line $AEC$ where $E=(1,1)$, by cutting off the pieces of length $r$ near $A$ and $C$. 

To prove minimality, let $\gamma$ be another shorter $r$-covering path. It must contain a point within distance $\sqrt{2}$ of $(0,0)$, call this point $Z$. As a calculus exercise, one can verify that $|AZ|+|CZ|\ge 2\sqrt{5}$, attained only if $Z=(1,1)$. (Or, instead of using calculus, recall that the locus of points of constant distance-sum from $A$ and $C$ is an ellipse  with $A$ and $C$ as foci.) Since $\gamma $ must also visit an $r$-neighborhood of $A$ and $r$-neighborhood of $C$, it follows that its length is at least $2(\sqrt{5}-r)$, as realized by $DEF$ on the sketch. (The angle $\angle DEF$ is greater than $120$ degrees, which implies that the union of $DE$ and $EF$ is the shortest continuum containing $D,E,F$, the Fermat point of this triangle.) 
