Planar compact connected set whose boundary has a finite length is arcwise connected

The answer is positive. First suppose that $K=\partial K$. I will show that in this case $K$ is the image of a curve. By definition of Hausdorff measure, for every $\epsilon>0$ we can cover $K$ with open disks of radii $\leq\epsilon$ so that the sum of the radii is at most $A$, where $A$ is independent of $\epsilon$. Since $K$ is compact, we can choose a finite subcovering with the same properties. The union of the disks of this subcovering is connected (otherwise $K$ will be disconnected). So by discarding if necessary some redundant disks we arrange that the union of the circles is connected.

Now consider this union of the circles as a graph (the vertices are intersection points of circles, the edges are the arcs into which these intersection points break the union of the circles). Notice that the degree of each vertex is even, since any number of circles with a common point will give even number of half-edges meeting at this point. Therefore, by Euler's theorem there is a closed path $\gamma$ passing through every edge exactly once. This path has length $\leq 2\pi A$.

Setting $\epsilon=1/n$, and parameterizing $\gamma$ by the the arclength, we obtain continuous functions $\gamma_n:[0,2\pi A]\to R^2$. This family is equicontinuous, since all functions satisfy the Lipschitz condition with constant $1$. Therefore by Arzela'a theorem there is a subsequence which converges uniformly to a curve $\gamma_\infty$. Since every point of $K$ is at a distance $\leq 1/n$ from $\gamma_n$, the image of $\gamma_\infty$ equals $K$, so we parameterized $K$ as a curve.

Now let us prove the general case. We know that every component of $\partial K$ is a rectifiable curve, and the sum of the lengths of these curves is finite. Let $x$ and $y$ be two points of $K$. Connect them with a segment $[x,y]$. If this segment is in $K$, we are done. If not, the intersection $[x,y]\cap cK$ consists of open intervals. We order these intervals into a sequence $(I_n)$ so that the lengths are non-increasing. Let $(a_1,b_1)$ be the longest of these intervals It belongs to a component $D$ of $cK$, and $\partial D$ is a closed connected set which belongs to $\partial K$. Therefore it belongs to some component of $\partial K$ which is a rectifiable curve. Replace this interval $(a_1,b_1)$ by a piece $\gamma_1$ of this curve. It is possible that $\gamma_1$ serves also some further intervals of our sequence $(I_n)$. Let $(a_2,b_2)$ be the first interval in the sequence $I_n$ whose endpoints are not on $\gamma_1$. Then repeat the procedure. In this way we obtain a sequence of curves $\gamma_k$ and the remainder of the original interval $[x,y]$ which together make a curve from $x$ to $y$, (even a rectifiable one).