Prove that $[7, 9]$ is covered by this family of intervals

Question

For each $x \in\mathbb{R}$ let $r_{x} > 0$. Suppose $r_x$ is unspecified and can be different with each $x$. Consider the uncountable family of open intervals:

$U = \{I_{r_x}(x): x \in\mathbb{R}\}$.

The interval $[7, 9]$ is covered by the intervals of $U$. Explain why there exists a finite number of the intervals of U which cover $[7, 9]$. What are those intervals, specifically?

EDIT: My solution: Suppose [7,9] is covered. Since [7,9] is closed and bounded, by the Heine-Borel Theorem, it follows that it is compact. Thus for any open cover, there exists a finite sub cover.

Answer 1

But the explicit assumption started with is that $[7,9]$ is covered by the intervals in $U.$ So if all had $x=100$ and $r_x=1$ they would not cover $[7,9],$ against the initial assumption.

Answer 2

Compactness guarantees you only need a finite number out of the collection to cover the closed interval. You know the interval is covered because for each $x$, $x\in I_{r_x}$.

Unless you're given some of the values of $r_x$ for a few given $x$, you can't really say which ones cover the interval specifically