I'll apply the following simple result:
THEOREM Let $f : I\rightarrow X$ be an arbitrary non-constant continuous function (a path) of interval $I:=[0;1]$ into an arbitrary topological space $X$. Then there exist continuous maps $\alpha:I\rightarrow I$ and $g:I\rightarrow X$ such that $f = g\circ \alpha$, and $g$ is not constant on any non-empty open subinterval of $I$.
Here is a simple positive solution for the question of this thread, and proof:
Let $\mathbb C := \mathbb R^2$ be the complex plane. Let $K \subseteq \mathbb C$ be a Knaster pseudo-arc. Let $$L := i\cdot K := \{i\cdot z : z \in K\}$$
where $i^2=-1$. Let $D$ be a dense countable subset of $\mathbb C$. Define $$A := \left(\bigcup_{d\in D}\ \left(d+K\right)\right)\cup\left(\bigcup_{d\in D}\ \left(d+L\right)\right)$$
where $d+X := \{d+x:x\in X\}$. Finally, let $$B := \mathbb C\setminus A$$
Then $\dim(B) = 0$, and $B$ does not contain any non-constant path.
Also, there does not exist any non-constant continuous map $f : I \rightarrow A$ --indeed, if there was one then we may assume that it is not constant on any open subinterval of $I$. Then the inverse images:
$(\bigcirc^{-1}f)(d+K)\quad$ and $\quad(\bigcirc^{-1}f)(d+L)$
would be 0-dimensional closed subsets of $I$, for every $d\in D$. Thus $I$ would be a countable union of 0-dimensional closed subsets, which is a contradiction. It means that $A$ does not contain any image of any non-constant path.would be 0-dimensional closed subsets of $I$, for every $d\in D$. Thus $I$ would be a countable union of 0-dimensional closed subsets, which is a contradiction. It means that $A$ does not contain any image of any non-constant path.
This completes a positive answer to the Question of this thread.
