This is not actually a question asked by me. But since I do not know the answer, I would love to know if someone here could answer it.



Is every locally connected subset of Euclidean space R^n locally path connected ?
No. There are locally connected subsets of $\mathbb{R}^2$ which are totally path disconnected. See my answer to this old MO question "[Can you explicitly write R2 as a disjoint union of two totally path disconnected sets?](https://mathoverflow.net/questions/156/can-you-explicitly-write-r2-as-a-disjoint-union-of-two-totally-path-disconnected)". Also, Gerald Edgar's response to the same question says that such sets cannot be totally disconnected, although he does not mention local connectedness. In fact, the sets given by my answer are locally connected, so provide a counterexample to your question.


As in the linked question: Let $S$ be a subset of the reals such that $S\cap[a,b]$ and $S^{\rm c}\cap[a,b]$ cannot be written as a countable union of closed sets for any $a < b$ (e.g., this [explicit example of a non-Borel set](https://planetmath.org/alebesguemeasurablebutnonborelset)). Then, $A=(S\times\mathbb{Q})\cup(S^{\rm c}\times\mathbb{Q}^{\rm c})$ and $B=(S\times\mathbb{Q}^{\rm c})\cup(S^{\rm c}\times\mathbb{Q})$ partition the plane into a pair of locally connected and totally pathwise disconnected sets.


That they are totally pathwise disconnected is proven in my answer to the linked question. Let us show that $A\cap U$ is connected for any nonempty 'open rectangle' $U=(x\_0,x\_1)\times(y\_0,y\_1)$. If not, there would be nonempty disjoint open sets $V,W\subset U$ with $A\cap(V\cup W)=A\cap U$. If $\pi(x,y)=x$ is the projection onto the *x*-axis then $\pi(V)\cup\pi(W)=\pi(V\cup W)=(x\_0,x\_1)$ is connected. So, we can find a nontrivial closed interval $[a,b]\subseteq\pi(V)\cap\pi(W)$.
Now, for every $x\in S^{\rm c}\cap[a,b]$, the line segment $\lbrace x\rbrace\times(y\_0,y\_1)$ intersects with both $V$ and $W$ and, by connectedness of line segments, it will intersect with $U\setminus(V\cup W)$. Hence, there is a $q\in\mathbb{R}$ with $(x,q)\in U\setminus(V\cup W)\subset B$. So, $q\in\mathbb{Q}$. For each rational $q$, let $S\_q$ be the (closed) set of $x\in[a,b]$ such that $(x,q)$ is in $U\setminus(V\cup W)$.
Then, $S^{\rm c}\cap[a,b]=\bigcup\_{q\in\mathbb{Q}}S\_q$ is a countable union of closed sets, giving the required contradiction. Hence $A$ is locally connected and, exchanging $S$ and $S^{\rm c}$ in the argument above, so is $B$.


