If you have Munkres (2nd ed.): Thm 25.4 states:
A space $X$ is locally path-connected iff for every open set $U$ of $X$, each path-component of $U$ is open in $X$.
So if indeed $U$ is open and connected it has (as all spaces) a decomposition into path-components, which are open in $X$ and thus open in $U$ too, and by being a partition, they are also closed (the complement is also a union of open sets) in $U$. So by connectedness there can be only one path-component.
25.5 even says (part 2 of it)
If $X$ is a topological space, and $X$ is locally path-connected, its components and path-components coincide.
Apply this to $X=U$ (which is locally path-connected as an open subspace of a locally path-connected space) and you're done right away.
Note, this is assuming you use the same definition of local path-connectedness as Munkres does (which is non-standard): every neighbourhood $U$ of $x$ contains a path-connected neighbourhood $V$ of $x$.
The definition in e.g. Engelking is:
for every open set $U$ and every $x \in U$ there is an open neighbourhood $V$ of $x$ such that for any $y \in V$ there is a path $p: [0,1] \to U$ connecting $x$ to $y$.
Note that $V$ is not supposed to be itself path-connected, as Munkres does. So the latter has a stronger notion, so maybe this fact only holds for the stronger notion; at least the proof does.