Path-connectedness, does the set need to be open? I have this exerise:

Let $X$ be locally path connected. Show that every connected open
  set in X is path-connected.

But does the set need to be open? I have this theorem in the chapter:

If $X$ is a topolocial space, each path component of X lies in a
  component of X. If X is locally path connected, then the components
  and the path components of X are the same.

Does not this theorem tell us directly that it holds for all subsets, not just open subsets?

Yes, the assumption that the set is open is necessary. The plane $\mathbb{R}^2$ is locally path-connected, and the "topologist's sine curve" $$S = \operatorname{closure} \bigl\{(x,\sin(1/x) \mid x > 0 \bigr\}$$ is connected (because it's the closure of a connected set), but it's not path-connected (a classical exercise). And since of course $S$ isn't a path component of $\mathbb{R}^2$, you cannot use the theorem you mention to say that $S$ is path-connected (and it isn't).
