Bisection of intervals is the right idea. First show that there exists a half open interval $\tilde I$ with $Q(\tilde I)=1$. This is true because $\mathbb R$ can be written as a countable disjoint union of such intervals, and $Q(\mathbb R)=1$, so due to $\sigma$-additivity, they can't all have measure $0$. Then define $I$ as the closure of $\tilde I$. Now construct a sequence of intervals $I_n$ by bisecting $I$ and choosing the half with measure $1$, and repeating iteratively. Always choose the intervals to be closed, so we can use the nested intervals theorem in the end. Now $I_n$ is a sequence of closed nested intervals whose length converges to $0$. Since $\mathbb R$ is complete, their intersection is a singleton set (nested intervals theorem), and due to $Q$'s continuity from above, the measure of that singleton is $1$. The one element inside it is $a$.