Showing that every connected open set in a locally path connected space is path connected

Solution 1:

That's a typical connectedness argument. As pointed out by muzzlator, fix a point $x$ in $\Omega$ your open connected subset. Then consider $\Omega_x$ the set of points in $\Omega$ which are path connected to $x$ within $\Omega$. This is nonempty as $x$ belongs to it. This is open in $\Omega$ by local path connectedness. And by local path connectedness again, it is easily seen that the complement $\Omega\setminus\Omega_x$ is open in $\Omega$. So $\Omega_x$ is a nonempty open/closed subset of $\Omega$. Thus $\Omega_x=\Omega$. Note that the fact that $\Omega$ is open is implicitly used in both steps.

Solution 2:

Here's an alternative way. Let $S$ be the set of all paths reachable from a point $x$. This is an open set due to the local path connectedness. Use this to form a disconnection of $X$ if $S$ wasn't the same as $X$.