Let $X$ be a locally path connected compact space. Show that $X$ has only finitely many path components.
Since $X$ is locally path connected each point $x \in X$ has a neighborhood $U_x$ such that there exists path connected $V_x$ for which $x \in V_x \subset U_x$.
Since $X$ is compact any open cover has a finite subcover.
Now if we assume that $X$ had infinitely many path components, then I think I should be able to contradict compactness, but I cannot figure out a way to show this. I only now that the path components parition $X$, but there is no further information about them. How can we get a contradiction here?
In a locally path connected space the path connected components are open. Thus the collection of the path connected components is an open cover of $X$ and from compactness of $X$ has a finite subcover. Now notice that the subcover has to be the cover itself as the path connected components are disjoint.