Edit: I've asked a followup question.
Suppose $X$ is a topological space such that any two points $x_0,x_1\in X$ are connected by an immersive path, i.e. there is a locally homeomorphic embedding $\gamma\colon [0,1]\to X$ such that $\gamma(0)=x_0$ and $\gamma(1)=x_1$.
Does it follow that $X$ is arc-connected, i.e. that we can choose $\gamma$ to be an actual homeomorphic embedding?
Note that we can assume without loss of generality that $X$ is the image if an immersive path, in particular that it is a compact space which is covered by (finitely many, by compactness) arcs. Further, the conclusion is true if $X$ is Hausdorff (in fact, just being path-connected is sufficient in this case).
(This is sort of motivated by my ruminations related to this question.)
This is a suggestion at an answer that is a bit too long to write up as a comment. Apologies if it is baloney. Let $A = [0, 1] \times \{0, 1\} \subseteq \Bbb{R}^2$ and take all sets of the form $(a, b) \times \{i\} \cup \{x\} \times \{ 1 - i\}$, where $0 \le a < x < b \le 1$ and $i \in \{0, 1\}$ as the subbasis of a topology on $A$. This is along the lines of the line with two origins, but now I am arranging for every point in the interior of the unit interval to have a doppelganger. Now let $B$ be $\{1\} \times [0, 1]$ equipped with the usual topology and take $X$ to be $A \cup B$ with the topology induced by the topologies given for $A$ and $B$. Then any two points in $X$ are connected by an immersive path (you can just ignore the doppelgangers in a neighbourhood of any point), but no point with a doppelganger is connected by an arc to its doppelganger (as the image of an arc must be Hausdorff, but no subset of $X$ containing a point and its doppelganger is Hausdorff).