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.
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.
Best Answer
Compactness of certain sets is not needed.
For the first part of your question you will find an answer in Definition of locally pathwise connected. But for the sake of completeness let us prove once more that the following are equivalent:
(1) $X$ is locally connected (locally path connected), i.e. has a base consisting of open connected (open path connected) sets.
(2) Components (path components) of open sets are open.
(1) $\Rightarrow$ (2): Let $\mathcal{B}$ be a base for $X$ consisting of open connected (open path connected) sets. Let $U \subset X$ be open and $C$ be a component (path component) of $X$. Consider $x \in C$. By assumption there exists $V \in \mathcal{B}$ such that $x \in V \subset U$. Since $V \cap C \ne \emptyset$, we see that $V \cup C$ is a connected (path connected) subset of $U$ which contains $C$. By definiton of $C$ we see that $V \cup C = C$, i.e. $V \subset C$. Hence $C = \bigcup_{V \in \mathcal{B}, V \subset C} V$. In particular, $C$ is open in $X$.
(2) $\Rightarrow$ (1): Let $U \subset X$ be open. For any $x \in U$ the component (path component) of $U$ containing $x$ is open, hence $U$ is the union of open connected (open path connected) sets.
Now, if $X$ is locally path connected, then it is also locally connected. Hence components and path components of open sets are open. Moreover, components and path components of open sets agree (this applies in particular to $X$ itself). To see this, consider an open $U \subset X$. Each path component $C$ of $U$ is contained in a component $C'$ of $U$. Assume $C \subsetneqq C'$. Let $C_\alpha$ be the path components of $C'$. They are again open, and we must have more than one. Then $C'$ can be decomposed as the disjoint union of two non-empty open sets (e.g. $C_{\alpha_0}$ and $\bigcup_{\alpha \ne \alpha_0} C_\alpha$). This means that $C'$ is not connected, a contradiction. We conclude $C = C'$.