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   $I$.

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Here is a simple positive solution for the question of this threa, and proof:

Let &nbsp; $\mathbb C := \mathbb R^2$ &nbsp; be the complex plane. Let &nbsp; $K \subseteq \mathbb C$ &nbsp; be a Knaster pseudo-arc. Let
$$L := i\cdot K := \{i\cdot z : z \in K\}$$

where &nbsp; $i^2=-1$. &nbsp; Let &nbsp; $D$ &nbsp; be a dense countable subset of &nbsp; $\mathbb C$. &nbsp; Define
$$A := \left(\bigcup_{d\in D}\ \left(d+K\right)\right)\cup\left(\bigcup_{d\in D}\ \left(d+L\right)\right)$$

where &nbsp; $d+X := \{d+x:x\in X\}$. &nbsp; Finally, let
$$B := \mathbb C\setminus A$$

Then &nbsp; $\dim(B) = 0$, and &nbsp; $B$ does not contain any non-constant path.

Also, there does not exist any non-constant continuous map &nbsp; $f : I \rightarrow A$ --**indeed**, if there was one then we may assume that it is not constant on any open subinterval of &nbsp; $I$. &nbsp; Then the inverse images:
<p>
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $(\bigcirc^{-1}f)(d+K)\quad$ &nbsp; and &nbsp; $\quad(\bigcirc^{-1}f)(d+L)$
</p>
would be 0-dimensional closed subsets of &nbsp; $I$, &nbsp; for every &nbsp; $d\in D$. Thus &nbsp; $I$ &nbsp; would be a countable union of 0-dimensional closed subsets, which is a contradiction. It means that &nbsp; $A$ &nbsp; does not contain any image of any non-constant path.

This completes a positive answer to the Question of this thread.