How to finish this proof (or sketch)?

I'm trying to prove that a manifold $M$, that is connected, is pathwise connected. I know the standard proof of this theorem: just use that the set of points that can be joined to a point $x \in M$ is clopen. However, I'm trying to find a more direct proof. I want to find explicitly a path that connect any two given points $x, y \in M$. Of course, I can create a sequence of points $(z_\alpha)_{\alpha \in A}$ indexed by some ordinal and consequently a sequence of sets $(U_\alpha)_{\alpha \in A}$ homeomorphic to the euclidean space, such that each open set $U_{z_{\alpha}}$ intersects with its successor and, then, concatenate each path to the other inside the intersection. However I don't know how to find a finite sequence (whether it's possible) nor to proof that this concatenation of paths will be continuous. Is there some way to finish it or an analogous proof?

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It's always a good idea to check with the long line. – Hagen von Eitzen Feb 15 '13 at 7:50
@HagenvonEitzen But the long line is compact, so I think the concatenation could become finite. – user40276 Feb 15 '13 at 7:53
@HagenvonEitzen Sorry, I thought it was the compactification. Ignore the previous comment. – user40276 Feb 15 '13 at 7:58

Here is my partial attempt, which eventually gives a constructive proof that connected and compact manifolds are path-connected.

Lemma 1: Any connected set that admits a finite open cover $\{U_i\}_{i = 0}^n$ of path-connected sets is itself path-connected.

Proof:

$U_0$ is path-connected.

For $n \geq 1$, assume by induction that $\bigcup_{i \in I}U_i$ is path-connected for any $I \subset \{0, 1, \dotsc, n\}$ (proper subset) such that $\bigcup_{i \in I}U_i$ is connected.

Suppose that $\bigcup_{i = 0}^nU_i$ is connected; let $\Gamma$ be the set of components of $(\bigcup_{i = 0}^nU_i) - U_0$; if $\Gamma = \emptyset$, then $\bigcup_{i = 0}^nU_i \subseteq U_0$ which is path-connected. Otherwise, choose some $C \in \Gamma$. Note that $$V = \bigcup\{U_i\ |\ i \in \{0, 1, \dotsc, n\}, U_i \cap C \neq \emptyset\}$$

is connected, and thus path-connected by the induction hypothesis ($U_0 \nsubseteq V$). By a theorem of Kuratowski's, $W = (\bigcup_{i = 0}^nU_i) - C$ is connected. Since $V$ and $W$ form an open cover of the connected $\bigcup_{i = 0}^nU_i$, they must intersect. It follows immediately that $\bigcup_{i = 0}^nU_i = V \cup W$ is path-connected.

In particular, since $V \cap W$ contains some $x$, we may concatenate paths with endpoint $x$ to explicitly construct paths, which is what the question wishes for. This construction can be inductively built up from cases with lower $n$. $\Box$

Corollary 1: Any connected manifold $M$ that admits a finitary atlas $\{(U_i, \phi_i\}_{i \in I}$ is path-connected.

Proof:

Apply Lemma 1 by noting that $\{U_i\}_{i \in I}$ is a finite open cover of the connected $M$, where each $U_i$ is homeomorphic to Euclidean space, which is path-connected. $\Box$

Proposition 1: Any connected and compact manifold $M$ is path-connected.

Proof:

Consider the atlas $\{(U_x, \phi_x)\}_{x \in M}$, where $U_x$ is some open neighborhood of $x$ that admits a homeomorphism $\phi_x$ to Euclidean space. $\{U_x\}_{x \in M}$ is an open cover of $M$, thus there exists a finite $I \subseteq M$ such that $\bigcup_{x \in I}U_x = M$. Apply Corollary 1 to the finitary atlas $\{(U_x, \phi_x)\}_{x \in I}$. $\Box$

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