Is separability + connectedness in metric spaces sufficient to guarantee path connectedness? Are these necessary conditions? Saw someone make this claim, but I'm not certain this is true. Can't think of counterexamples in either direction. Any help would be appreciated.
 A: The Topologist's sine curve is a classical counterexample.
A: Complementing mathcounterexamples.net's answer, it's not a necessary condition either. Consider an "uncountable starfish" - take uncountably many copies of the unit interval, and glue them together at their $0$-endpoints. This is path-connected but not separable.

A bit more rigorously: take an uncountable set $A$, and consider the set $X=(A\times(0,1])\sqcup\{*\}$. We put a metric $d$ on $X$ as follows:

*

*For each $x\in X$ we set $d(x,x)=0$.


*For each $a\in A, i\in (0,1]$ we set $d(*,\langle a,i\rangle)=d(\langle a,i\rangle, *)=i.$


*For each $a\in A$ and each distinct $i,j\in (0,1]$ we set $d(\langle a,i\rangle , \langle a,j\rangle)=\vert i=j\vert$.


*Finally, for each distinct $a,b\in A$ and each $i,j\in (0,1]$ we set $d(\langle a,i\rangle ,\langle b,j\rangle)=i+j.$
It's a good exercise to check that this $d$ is in fact a metric on $X$, and that the resulting topology is path-connected. But it's clearly not separable: given a countable set of points $C\subseteq X$, since $A$ is uncountable we can find some $a\in A$ which is not the left coordinate of any point in $C$, and then $\langle a,1\rangle$ is at distance at least one from any point in $C$.
