TITLE: Is every locally connected subset of Euclidean space R^n locally path connected ?
QUESTION [21 upvotes]: 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.

REPLY [4 votes]: An easy counterexample:
The basic space that is connected but not locally connected is the closure of the graph $y=\sin(1/x)$ ("the topologist sine curve").
Call it $S$. Pick a countable set $D$ on the graph $y=\sin(1/x)$ whose closure is $D\cup \{0\}\times [-1,1]$. For each point $z \in D$ add a copy $S_z$ of $S$ (in $R^3$) starting from $z$ and ending on an interval in $0 \times [-1,1]$ so that the diameter of $S_z$ is at most twice the distance from $z$ to the  $y$-axis and $S_z$ intersects $S_w$ only along the $y$-axis if $z$ is different from $w$. The union $C$ of $S$ and all $S_z$, $z \in D$, is clearly locally connected.
Any arc from $w$ in $D$ to the $y$-axis contained in $C$ would have to be contained in $S$ (it intersects each $S_z$ at most in $z$), a contradiction.
Added after reading comments:
Let's pick a sequence of planes $\Pi_n$ containing the $y$-axis and converging from above (from the point of view of the half-plane $x\ge 0$) to the $xy$-plane where $S$ resides. Enumerate points $z\in D$ as $z_n$, $n\ge 1$.
From each $z_n$ go up until you hit $\Pi_n$ and then create a scaled copy of $S$ lying on $\Pi_n$ and ending on the $y$-axis.
That should guarantee $S_z\cap S_w \subset y-\text{axis}$.
Why is $C$ locally connected?
The issue is only with point on the $y$-axis. Given any $\epsilon > 0$ and $-1\leq y\leq 1$, consider the union of all $S_z$ that intersect $0\times [y-\epsilon,y+\epsilon]\times 0$ and are of diameter less that $\epsilon$. Add $\epsilon$-neighborhoods of end-points of $S_z$s in $S$. That set is connected, of small diameter, and it contains $(0,y,0)$ in its interior rel.C.