Of course if the set was arbitrary it's easy considering Q to find a solution. But is there an example of a open set? I believe the answer is no.
My attempt: By Classification of open sets in R, the set must be a countable union of disjoint open intervals. Thus it's closure is the closure of that union. Which is the union of the closure of those intervals. Since the measure of intervals does not change with the addition of endpoints, the total measures cannot change.
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Consider the set of dyactic fractions $\left\{\frac{a}{2^m}\right\}$ where $a$ is an odd integer and $m$ is a positive integer. This is a dense set.
Now consider the union of intervals $$\bigcup_{a,m} \left(\frac{a}{2^m}-\frac{k}{2^{2m}}, \frac{a}{2^m}+\frac{k}{2^{2m}} \right)$$ for some real constant $k$ with $0 \lt k \lt 2$. Since this is a union of open intervals, it is also an open set. And it does not include $0$ or $1$ so its intersection with $[0,1]$ is also an open set. Now consider that intersection.
The intersection with the unit interval has positive measure, and for some $k \approx 0.55985$ it has measure $\frac12$, while its closure is the unit interval with measure $1$.
The following chart, taken from something I did $15$ years ago shows how the measure of the boundary of the open set with $k$, i.e. $1$ minus the measure of the open set. It is a strictly decreasing continuous function which has a dense set of points where its derivative is zero.
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Say $(r_n)$ is an enumeration of the rationals in $(0,1)$. Choose $a_n>0$ with $\sum 2a_n<1/2$ and let $$E=(0,1)\cap\bigcup_n(r_n-1_n,r_n+1_n).$$Then $E$ is open, dense, and $m(E)<1/2$.
Now for $\alpha\in(0,1)$ let $$S_\alpha=E\cup(0,\alpha).$$Note that $m(S_\alpha)$ depends continuously on $\alpha$; hence there exists $\alpha$ with $m(S_\alpha)=1/2$.
Enumerate $\Bbb Q\cap[0,1]$ as $q_1,q_2,q_3\ldots$. Let $$ U=\bigcup_{n\in \Bbb N}(q_n-2^{-n-2},q+2^{-n-2}).$$ Then $$\mu(U)\le\sum_{n=1}^\infty 2^{-n-1}=\frac12 $$ but $\overline U=[0,1]$.