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I am following D.gale's outline of proof for the classification theorem.
(http://www.igt.uni-stuttgart.de/eiserm/lehre/2014/Topologie/Gale%20-%201-manifolds.pdf)

For two charts $(U,\phi)$ and $(V,\psi)$ of a 1-manifold $X$ (which meets but not included in each other),
I have these two propositions
1. Suppose $X$ is compact and connected.
if $X=U\cup V$ and $U \cap V$ has two connected components, then $X$ is homeomorphic to a circle $C^1$.
2. If $X=U \cap V$ is connected, $U \cup V$ is homeomorphic to $\mathbb R$.
(Plus, It is also given that $U \cap V$ has at most two components.)

I am trying to prove that every compact connected 1-manifold $X$ is homeomorphic to $C^1$ by generalizing the first proposition.
To be concrete, I took a finite open cover $\mathcal O$ of $X$ by its subsets which are homeomorphic to $\mathbb R$ which has no proper subcover.
As Gale suggested in his text, I tried to use induction on $\vert \mathcal O \vert$.
Proposition 1 covers the case where $\vert \mathcal O \vert = 2$, and I am pretty sure that I would have to use Proposition 2 to make the $k \to k+1$ argument.
However, I cannot show that there is at least one pair $(U_i,U_j)$ of elements in $\mathcal O$ such that $U_i \cap U_j$ is nonempty and connected, which seems necessary to apply the Propositon 2 to this argument.
How can I get around with this problem?
Thanks in advance.

Tl;dr : Help me in showing the compact case of classification theorem for 1-manifolds without boundary, from the two propositions mentioned above.
(edit : typo)

Spiche
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    If $U_i \cap U_j$ is non-empty but connected, the $U_i \cup U_j$ is homeomorphic to $S^1$. Then since $U_i \cup U_j$ is open and closed, $X \cong S^1$. If all $U_i \cap U_j$ are empty, then $X$ is not connected as it is the disjoint union of proper open subsets. – PVAL-inactive Jul 13 '17 at 00:00
  • I thought $U_i \cup U_j$ is $S^1$ when $U_i \cap U_j$ has two components. Also, I don't understand your second sentence. Could you explain it more explicitly? – Spiche Jul 13 '17 at 00:10
  • In the first sentence of my last comment there is a typo. It should be "If $U_i \cap U_j$ is non-empty but not-connected,..." – PVAL-inactive Jul 13 '17 at 00:23
  • @PVAL-inactive No. never mind. I was super dumb not to understand your word. It is way simpler than my expectation! Thanks a lot :) – Spiche Jul 13 '17 at 01:01

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