Let $M$ be a subset of a metric space $(E,d)$. I have just proved that $M$ path-connectedness implies $M$ connectedness. Now I need to show that if $M$ is connected and every open ball in $E$ is path-connected (in $E$), then $M$ is path-connected. I don't really know where to begin .Can you give me a hint?
EDIT: I'm really really confused. OK, given $x\in S$ I have to prove that the set $ S =\{y\in M:$ there exists a path in $S$ between $x$ and $y \} $ is open and closed ( it's trivially non empty, because $x\in S$). I have been trying to prove it's open, I have to prove that for an arbitrary $s\in S$ there is a ball $B$ centered in $s$ such that $B\bigcap M$ is contained in $S$. However, although every $B$ is path-connected I cannot assure for any $B$ that $B\bigcap M$ is path-connected , which is what I need. Are you sure I can do it with these hypothesis? Please, I really need help with this.
EDIT2: Do we need $M$ to be an open set?
EDIT3: Tell me if I'm wrong, but I think the topologist's sine is a counterexample for the case when $M$ is not open. Your answers are OK assuming $M$ is open
We have to assume that $M$ is open or that every open ball in $M$ is path connected in $M$. Otherwise, as the OP himself pointed out, the topologist's sine curve provides a counterexample. Assuming that the hypothesis is modified here is a hint: let $x \in M$. Let $S$ be the set of all points $y \in M$ such that there is a path in $M$ from $x$ to $y$. It is fairly elementary to verify (using path connectedness of open balls) that $S$ is open and closed in $M$. Since $M$ is connected (and $S$ is not empty) it must be equal to $M$.