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?
 A: 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).
A: The "topologist's sine curve" is a nice example. Wouldn't you rather see a nonconstructive horror?
Assuming the axiom of choice, there is a "Bernstein decomposition" of the plane, i.e., $\mathbb R^2$ is the union of two disjoint sets $B_1$ and $B_2,$ each of which has nonempty intersection with every uncountable closed subset of $\mathbb R^2.$ These sets $B_i$ are famously not Lebesgue measurable, but what's pertinent to your question is that they are both connected sets, and that neither of them contains any nonconstant path.
