Let $M$ compact and $\mathcal{U}$ an open covering of $M$ that satisfies: each $p \in M$ is contained in at least two members of $\mathcal{U}$. Show that $\mathcal{U}$ has a finite subcovering with the same property.
Writing $\mathcal{U} = \{U_{i}\}$, since $M$ is compact, we have
$$M \subset U_{\alpha_{1}} \cup \cdots \cup U_{\alpha_{k}}.$$
How to ensure that taken $p \in M$, $p \in U_{\alpha_{n}}\cap U_{\alpha_{m}}$? Intuitively, could it not be $p \in U_{\alpha_{1}}\cap U_{\alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?
Hint: Let $\mathcal U = \{ U_i : i\in I\}$ be the original open covering of $M$.
Can you show that $\mathcal V = \{ U_i \cap U_j : i, j \in I, i \neq j \}$ is also an open covering of $M$?
Then use the compactness of $M$ to find a finite refinement of $\mathcal V$...