 Solution

Take the topologist's sine curveIt is closed and bounded, hence compact. The oscillating graph is connected and accumulates on every point of the vertical segment, so its closure $S$ is connected. No path joins the graph to the vertical segment, as continuity of the first coordinate and the unbounded oscillation of $\sin (1/x)$ would fail at the first time the path reaches $x=0$. Thus $S$ is not path connected.