The Topologist's Sine Curve is well-known:
The set $S = \{(0,0)\} \cup \{(x, \sin(1/x))\ |\ x \in \left]0,1\right]\}$, as a subspace of $\mathbb{R}^2$, is connected but not path-connected.
An intuitive reason is that no path from $S - \{(0,0)\}$ can reach $(0,0)$ in a finite amount of time. However, what if I shrink the distances involved? That is, consider $$ S' = \{(0,0)\} \cup \{(x, x\sin(1/x))\ |\ x \in \left]0,1\right]\}. $$
Is $S'$ path-connected?
Let $$f:[0,1]\to\Bbb R:x\mapsto\begin{cases}x\sin\frac1x,&\text{if }0<x\le 1\\\\0,&\text{if }x=0\;;\end{cases}$$
then $f$ itself is already a path connecting any two points of $S'$.