Let $\displaystyle |U_\alpha|_{\alpha{\in}I}$ be a collection of open sets such that $\displaystyle [0,1]\subset{\bigcup_{\alpha{\in}I}}U_\alpha$

Prove $\displaystyle \exists$ a finite number of sets $\displaystyle U_{\alpha_{1}},...U_{\alpha_{n}}$ such that

$\displaystyle [0,1]\subset{\bigcup^n_{i=1}}U_{\alpha_{i}} $

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Clarification

The statement $\displaystyle [0,1]\subset{\bigcup_{\alpha{\in}I}}U_\alpha$ means $\displaystyle \exists$ an open set containing number $\displaystyle 0$, another (or the same set) containing number $\displaystyle 1$ and for every $\displaystyle x, 0\leq{x}\leq{1}, \exists U_\alpha$ such that $\displaystyle x\in{U_\alpha}$, so this implies $\displaystyle [0,1]\subset{\bigcup_{\alpha{\in}I}}U_\alpha$?