Are topologically well-behaved measure 0 subsets of $\Bbb R^2$ finite graphs?

Question

Conjecture: If $X\subseteq \Bbb R^2$ is locally simply connected (hence locally path connected), compact and Lebesgue measure $0$ then $X$ is homeomorphic to a finite graph.

It is clear that losing any of these hypotheses makes the conjecture false (e.g. Hawaiian earring for locally simply connected). Moreover the higher dimensional analogue,

$X\subseteq \Bbb R^n$ locally simply connected, compact, measure $0$, then $X$ is homeomorphic to a finite CW-complex of at most dimension $n-1$.

is false. Take the Hawaiian earring $H$ and embed it into $\Bbb R^3$ so that the convex disks bounded by the circles in $H$only intersect at the point where the circles intersect. Now just consider the union of $H$ and theses disks. The resulting space is a counterexample.

I really have no idea how to try and prove such a result if it is true (which I suspect), and the question seems relative resilient against standard topological counter-examples that I know.

Answer 1

Counterexample:$$X=([0,1]\times\{0\})\cup\bigcup_{n=1}^\infty(\{\frac1n\}\times[0,\frac1n])$$

Answer 2

Well, I found a counter-example... Take $H$ and delete an open arc from each circle (not hitting the point where the circles intersect).