"Redundant" finite subcovering of a compact space. Let $M$ be compact and $\mathcal{U}$ an open covering of M such that each $p \in M$ is contained in at least two members of $\mathcal{U}$. Show that $\mathcal{U}$ reduces to a finite subcovering with the same property.
I've been struggling with the question for quite a while but did not get anywhere..
 A: First, take a finite cover of your space $M.$
Note that since it is finite, the set of point $p \in M$ such that $p$ is only in one of your $\mathcal{U_i}$ will be a closed set. (You need to use the finiteness of your cover here)
Since this is closed, it will be compact (here I am assuming that your space is Hausdorff though). Then consider a cover of that set by elements of $\mathcal{U}$ that are not in your first cover.
A: Hint: move from $\mathcal{U}$ to another cover.
A: I would like to elaborate on part of Maxime Scott's solution given above, where it is told that the set $S$ of points which belong to only one member of the finite subcover has to be closed:
Call the finite subcover of $\mathcal{U}$ as $\mathcal{U}'$. The set of points belonging to at least two $U_i \in \mathcal{U}'$ is $$M\setminus S=\bigcup_{1\le i < j \le \lvert \mathcal{U}' \rvert} \{U_i \cap U_j : U_i \in \mathcal{U}'\}.$$ Since all $U_i\cap U_j$ are open, their union is open as well, so $M\setminus S$ is open. Hence $S$ is closed.
