In a locally path connected space, every open connected subset is path connected I know this has a solution here: 
Showing that every connected open set in a locally path connected space is path connected
But I would like to see if there is a fault in this simpler proof:
Let $C \subset X$ be open and connected. I know that every path component of $C$ is open in $X$ by the fact that $X$ is locally path connected. Since the path components of $C$ form a disjoint partition of $C$, and since they are open in $X$, and since $C$ is non empty, $C$ must equal one and not more of those path components. So $C$ is path connected.
 A: 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.
A: Your proof is (almost) right. Besides the fact that you should state that locally path connectedness implies that path components of $C$ are open in $X,$ you used another fact that I think should be stated explicitly in your proof to be complete: the fact that $C$ is open in $X.$ Every path component of $C$ is open in $X.$ Since $C$ is open in $X,$ every path component of $C$ is open in $C$ as well. Then each nonempty path component of $C$ is both open and closed in $C,$ and connectedness of $C$ implies that $C$ has only one path component: itself. 
