Is It Always Possible to Draw A Connected Compact Set in $\mathbb R^2$?

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 $S$.