First of all, your open set $(a,b)$ contains a closed interval $[c,d]$ with $c \lt d$. By its definition Lebesgue outer measure is monotone, hence it suffices to show that $\mu^{\ast}[c,d] \gt 0$.
I agree with William that induction won't help much. leo's argument is fine, but I prefer a more informal presentation of the argument to see what is going on.
Every cover of $[c,d]$ contains a finite subcover by compactness and dropping superfluous intervals only reduces the total length of the cover, so we may assume that we deal with finitely many intervals to begin with.
So, let $[c,d] \subset I_1 \cup \cdots \cup I_n$. Since $c$ is covered, we must have an interval $(a_1,b_1)$ in our family $\{I_1,\ldots,I_n\}$ with $c \in (a_1,b_1)$, or, equivalently, $a_1 \lt c \lt b_1$. If $b_1 \lt d$ then $b_1$ is covered by another interval $(a_2,b_2)$ in the family $\{I_1,\ldots,I_n\}$, so $a_2 \lt b_1 \lt b_2$. If $b_2 \lt d$ then $b_2$ is covered$\ldots$
Since we started with a finite family of intervals that covers $[c,d]$ we must at some point arrive at an interval $(a_j,b_j)$ with $a_j \lt d \lt b_j$ and we stop there.
In this way we produce a sequence $(a_1,b_1), (a_2,b_2), \ldots, (a_j,b_j)$ of intervals covering $[c,d]$ such that $a_{i+1} \lt b_{i}$ for $i = 1,\ldots,j-1$ and $a_1 \lt c$ as well as $d \lt b_j$. Now we can estimate
$$\begin{align*}
\sum_{i=1}^{n} |I_i| & \geq \sum_{i=1}^{j} (b_i - a_i) =(b_j - a_j) + (b_{j-1} - a_{j-1}) + \cdots + (b_1 - a_1) \\
&=
b_j + \underbrace{(b_{j-1} - a_j)}_{\geq 0} + \underbrace{(b_{j-2} - a_{j-1})}_{\geq 0} + \cdots + \underbrace{(b_1 - a_2)}_{\geq 0} - a_1 \\
&\geq b_j - a_1 \geq d-c.
\end{align*}$$
This works with an arbitrary finite family of intervals $\{I_1, \ldots,I_n\}$ covering $[c,d]$. Hence this shows that Lebesgue outer measure of $[c,d]$ is at least $\mu^\ast [c,d] \geq d-c$ and since it is clear that it is at most $d-c$, we have $\mu^\ast [c,d] = d-c$.
Finally, we should combine the above with what Carl said, namely that the outer measure of $[a,b]$ is equal to the outer measure of $(a,b)$ and we're done.
Ok, you have a collection $[a_1, b_1],...,[a_n, b_n]$ that covers $A$ of total diameter less than $\frac{\epsilon}{2}$. You know that the rationals in $[0,1]$ are measure $0$, so you can find $[c_1, d_1], ..., [c_m, d_m]$ covering $\mathbb{Q} \cap [0,1]$ of diameter less than $\frac{\epsilon}{2}$. Then together, these intervals form a cover of $[0,1]$ of total diameter less than $\epsilon$ which is a contradiction.
Best Answer
Your contrapositive seems to be wrong in two ways. An interval not being closed $[a,b]$ does not mean it is open, and the negation of "not of measure zero" should be "of measure zero". In any case the statement is missing the assumption $a<b$, at least for the Lebesgue measure.