# Finite subsets of $\mathcal{U}$ don't cover $C \setminus \{p\}$

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 ($a<b$), 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.)

-

-