This is a follow-up to this question: A continuous path between shapes .
Let $A$ and $B$ be two measureable, bounded, connected subsets of $\mathbb{R}^2$ such that $A\subseteq B$.
Does there exist a continuous path $f$ from $A$ to $B$ (as defined in that question), such that for every $t \in [0,1]$, $f(t)$ is connected?
Here are my attempts to prove that such a path exists:
1. Physical "proof"
Assume that the set $A$ is the mouth of a water-pipe and the set $B$ is a water-pool. The water starts flowing through the pipe at time 0. Initially, there is water only at the mouth $A$. As time advances, the body of water grows, until at time 1 the entire pool $B$ is filled with water. From physical considerations, the body of water is monotonically increasing, its area grows continuously, and it is always connected.
Do you have any idea, how to convert this physical description to a valid mathematical proof? (See a similar question in this MathOverflow thread).
2. Mapping proof
First, assume $B$ is simply-connected. Use Riemann's mapping theorem to find a map m from $B$ onto the disc with radius 2 centered at the origin. Call this disc $B'$ and call the image of $A$ under that mapping, $A'$. Let $f'$ be the path from $A'$ to $B'$ described in Christopher's answer. Because B' is convex, $f'(t)$ is always connected (is it?). Now, map $f'$ back to the original shapes (e.g. let $f = m^{-1}\circ f$). Riemann's mapping preserves continuity and connectedness (does it?), so the function $f$ is indeed continuous and $f(t)$ is connected.
If $B$ is not simply-connected, it can be converted into a simply-connected shape using the construction in Joseph O'Rourke's answer.
Can this sketch be converted into a full proof?
Here is a way to formalize your first idea, assuming $B$ is compact.
Consider $B$ as a metric space where the distance between two points, $d(x,y)$, is given by the length of the shortest path within $B$ joining $x$ and $y$. (This is just the standard definition of distance on a compact Riemannian manifold.) Then a ball of radius $r$ around a point $x$ is always connected, because any point $y$ in the ball is connected to $x$ by the path of length $\le r$ that determines $d(x,y)$.
Now apply user73985's answer, except replacing the Euclidean disk $C_{ct}$ with the analogous ball in this metric space.