Let X be locally path connected. Show that every connected open set in X is path connected.
The Attempt at a Solution
Let U be a connected open subset of X. Since, X is path connected, for any x in X and any neighborhood N of X, there exists a path connected neighborhood Vx of x contained in N. For any x in U, pick a basis element containing x, contained in U, and for that basis element, pick the path connected neighbourhood Vx contained in this basis element. Then U can be written as the union of the Vx's, where x is in U.
Now, since U is connected, this union must have a nonempty intersection. Since if this would not hold, we could pick any Vx, and then this Vx and its complement in U would be two nonempty disjoint and open sets whose union is U, contradicting the fact that U is connected.
Let x, y be any elements of U. Then x is in Vx and y in Vy. Take an element z from the intersection of Vx and Vy. Let f1: [a, b] --> Vx be the path from x to z, and f2 : [c, d] --> Vy the path from z to y. Let [b, d] be some segment homeomorphic to [c, d]. Then, by the pasting lemma, the function f : [a, d] --> Vx U Vy (given with f = f1 is t is in [a, b] and f = f2 if t is in [b, d]) is continuous, and hence a path from x to y.
So, U is path connected.
Sorry if this is a bit long, but I tried to be precise.