The answer is no. Let $\alpha$ be an irrational in the interval $[0,1]$, say $\alpha=1/\sqrt{2}$.
Look at the set of all open intervals of the form $(-1, \alpha -\frac{1}{n})$ or $(\alpha+\frac{1}{n},2)$, where $n$ ranges over the positive integers. Every rational in $[0,1]$ is in one of these intervals. However, this set of intervals cannot be replaced by a finite subset, since a finite subset would miss all the rationals that are close enough to $\alpha$.
Remark: We can replace the intervals $(-1,\alpha-\frac{1}{n})$ by intervals of the shape $(-1, a_n)$, where $(a_n)$ is a sequence of rationals that approaches $\alpha$ from below, and deal similarly with the $(\alpha+\frac{1}{n},2)$. Thus, in your question, replacing reals by rationals does not change the answer.