Is It Always Possible to Draw A Connected Compact Set in $\mathbb R^2$? Inspired by this answer, I wondered whether a printer could render all continuous functions "well enough". In particular, I am curious about the following statement:

Let $S$ be a compact, connected subset of $\mathbb R^2$. For any $\varepsilon>0$ define $$S_{\varepsilon}=\bigcup_{s\in S}B(s,\varepsilon)$$ where $B(s,\varepsilon)$ ithe ball of radius $\varepsilon$ around $s$ (under the usual metric) - that is $S_{\varepsilon}$ is $S$ "expanded" everywhere by $\varepsilon$. For all $S$ and $\varepsilon$, must there exist a curve $\gamma:\mathbb [0,1]\rightarrow\mathbb R^2$ of finite length such that $$S_{\varepsilon}=\bigcup_{x\in[0,1]}B(\gamma(x),\varepsilon)$$

Or, putting it informally:

Given a compact, connected set, is it possible for a printer, which always draws a swath of radius $\varepsilon$, to render the set as closely as possible?

My thinking is "yes", because, we can clearly do this if we allow any sloppiness -that is, if we don't care whether the the points within some $0<\varepsilon'$ of the boundary of $S_{\varepsilon}$ are covered. Hopefully, complex figures on the boundary will be swallowed up when we expand out the set $S$ - but I'm suspicious of pathological examples and can't seem to draw up a proof or counterexample.

A statement I believe is equivalent (and of which equivalence I believe a simple proof likely exists) is the following:

Define $F_{\varepsilon}(S)$ to be the "inflation" of $S$ by $\varepsilon$ - the set of points within $\varepsilon$ of some $s\in S$. Define $f_{\varepsilon}(S)$ to be the "deflation" of $S$ by $\varepsilon$ - the set of points $s\in S$ such that the ball of radius $\varepsilon$ around is contained in $S$. Then, we wish to show that the boundary of $f_{\varepsilon}(F_{\varepsilon}(S))$ has finite length for any compact $S$.

 A: Here's a proof that the boundary of $f_\epsilon(F_\epsilon(S))$ does have finite length for each $\epsilon > 0$. I'll use the following inequality for any compact $S\subseteq\mathbb{R}^2$,
\begin{align}
{\rm perimeter}(F_\epsilon(S))\le \frac2\epsilon{\rm area}(F_\epsilon(S)).&&{\rm(1)}
\end{align}
More on why this holds in a bit. First, I'll use it to prove the result asked for. Note that $f_\epsilon(S)$ is the set of points of distance at least distance $\epsilon$ from the complement of $S$ and, hence,
$$
\partial f_\epsilon(S)=\partial F_\epsilon(\mathbb{R}^2\setminus S),
$$
where $\partial S$ denotes the boundary of a set $S$.
So,
$$
\partial f_\epsilon(F_\epsilon(S))=\partial F_\epsilon(\mathbb{R}^2\setminus F_\epsilon(S)).
$$
Now, if $S$ is contained in the closed ball $\bar B_R$ of radius $R$ about the origin, then $F_\epsilon(S)$ is contained in $\bar B_{R+\epsilon}$ giving,
$$
\partial F_\epsilon(\bar B_{R+\epsilon}\setminus F_\epsilon(S))=\partial\bar B_{R+2\epsilon}\cup\partial F_\epsilon(\mathbb{R}^2\setminus F_\epsilon(S))=\partial\bar B_{R+2\epsilon}\cup\partial f_\epsilon(F_\epsilon(S)).
$$
Putting this together with (1),
\begin{align}
{\rm perimeter}(f_\epsilon(F_\epsilon(S)))&={\rm perimeter}(F_\epsilon(\bar B_{R+\epsilon}\setminus F_\epsilon(S)))-{\rm perimeter}(\bar B_{R+2\epsilon})\\
&\le\frac2\epsilon{\rm area}(F_\epsilon(\bar B_{R+\epsilon}))\\
&=\frac2\epsilon\pi(R+2\epsilon)^2 < \infty,
\end{align}
which concludes the proof.

Let me now show why (1) holds. There is a quick proof for $S$ a finite set, as given by Theorem 5.3 in the PhD thesis by Zoltán Gyenes. By compactness, for any $\delta > 0$, there exists a finite subset $A$ of $S$ such that $F_{\epsilon}(A)\supset F_{\epsilon-\delta}(S)$, so the boundary of $F_{\epsilon}(S)$ lies within a distance $\delta$ of the boundary of $F_\epsilon(A)$. Fixing $N$ points around the boundary of $F_\epsilon(S)$, these points all lie within $\delta$ of points on the boundary of $F_\epsilon(A)$ so, the length of the piecewise linear curve interpolating these points is bounded by
\begin{align}
N(2\delta)+{\rm perimeter}F_\epsilon(A)&\le 2 N\delta+\frac2{\epsilon}{\rm area}F_\epsilon(A)\\
&\le 2 N\delta+\frac2\epsilon{\rm area}F_\epsilon(S)
\end{align}
Letting $N$ go to infinity gives (1) for the compact set $S$.
