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I'm familiar with this concept and it makes sense, but the proof for it is eluding me.

Let $p \in C$ and consider the set:

$\mathcal{U}=\{\operatorname{ext}(a,b)\mid p\in (a,b)\}$

Therefore no finite subset of $\mathcal{U}$ covers $C \setminus \{p\}$.

It makes sense that a finite number of exteriors will never cover the continuum $C$ ($C$ being nonempty, having no first or last point, ordered ($a), and connected) without $p$ since $p$ will be in exactly none of the subsets $\mathcal{U}' \subset \mathcal{U}$. I'm not sure if that's enough (or maybe even true) though. If anyone can point me in the right direction it will be very much appreciated. (Also, note that $\operatorname{ext}(a,b)$ is the same as $C \setminus [a,b]$. This clarification is simply based on notation.)

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    I edited it. Thank you very much for the suggestion.2012-11-07

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$\newcommand{\ms}{\mathscr}\newcommand{\ext}{\operatorname{ext}}$Let $\ms{F}$ be a finite subset of $\ms U$, say $\ms F=\{\ext(a_1,b_1),\dots,\ext(a_n,b_n)\}$. Let $a=\max\{a_1,\dots,a_n\}$ and $b=\min\{b_1,\dots,b_n\}$; by hypothesis $a. To show that $\ms F$ does not cover $C$, it suffices to show that $(a,b)\ne\{p\}$, i.e., that at least one of the open intervals $(a,p)$ and $(p,b)$ is non-empty. In fact both are non-empty, and you can show this simply by showing that

$\qquad\qquad\qquad\qquad\qquad$if $x,y\in C$ and $x, then $(x,y)\ne\varnothing$.

This follows from the fact that $C$ is connected, though the details of the argument depend on just what definition of connectedness you’re using.