I came up with a proof for the title, and I can't figure out what is wrong with the proof.
First, we had a proposition proved in class:
Proposition (continuity along decreasing chain): Let $(X,\mathcal{M},\mu)$ be a measure space. Suppose $(C_n)_{n=1}^\infty$ are in $\mathcal{M}$ such that $C_1\supseteq C_2\supseteq \cdots$ and $\mu(C_1) <\infty$. Then, $\mu(\bigcap_{n=1}^\infty C_n) = \lim_{n\to\infty}\mu(C_n)$.
So let $C_n = \bigcup_{r\in(\mathbb{Q}\cap[0,1])} B(r,\frac1{2^n})$, i.e. union of open balls around rationals with radius $\frac1{2^n}$. Since $\mathbb{Q}$ is dense, then $C_n = [0,1]$, and $\mu(C_n) = 1$ for all $n\in\mathbb{N}$. Then, $\lim_{n\to\infty}\mu(C_n) = 1$. But $\bigcap_{n=1}^\infty C_n = \mathbb{Q}\cap[0,1]$ which must have measure zero. It seems contradict to the proposition?
Let $\Bbb Q\cap [0,1]=\{r_n:n\in \Bbb N\}.$ For any $\epsilon>0$ let $D(\epsilon)=\cup_{n\in \Bbb N}(-2^{(-n-1)}\cdot\epsilon+r_n,\,2^{(-n-1)}\cdot\epsilon+r_n).$
Then $\Bbb Q\cap [0,1]\subset D(\epsilon).$ So for the Lebesgue outer measure $\mu^o$ we have $$\mu^o(\Bbb Q\cap [0,1])\le \mu^0(D(\epsilon))\le \sum_{n\in \Bbb N}2\cdot 2^{-n-1}\epsilon=\epsilon.$$