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.

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 R^{2} as a disjoint union of two totally path disconnected sets?". 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 nonBorel set). 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 xaxis 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 $B$. 


Isn't every connected subset of ${\mathbb R}^n$ locally a Peano space? Then the answer would be "yes". Edit I assumed the subspace to be closed. Then it would be locally compact, hence locally Peano. I guess the main difficulty is for not closed subsets. 


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 halfplane $x\ge 0$) 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 endpoints of $S_z$s in $S$. That set is connected, of small diameter, and it contains $(0,y,0)$ in its interior rel.C. 


A partial answer is that this is true for compact sets (as stated in the answer by Mark Sapir) and also for all open sets $M$, because one can prove rather easily, that the set of all points of $M$ which can be joined to some fixed $x_0\in M$ is open and closed at the same time. 

