This post will prove an upper bound of $1/(4-2 \sqrt{2}) \approx 0.854$, improving slightly on Christian Remlings $0.87$. 

Consider $M < N < 2M$. At time $M$, there are $M$ intervals, each of length $\geq \mu/M$. Let $S$ be the number of those intervals which are subdivided at time $N$, we have $S \leq N-M$. So $M-S$ intervals remain undivided. We have
$$(M-S) \frac{\mu}{M} + (N-M+S) \frac{\mu}{N} \leq 1.$$
Letting $S$ go all the way up to $N-M$ makes $\mu$ as large as possible; in this case we have
$$(2M-N)\frac{\mu}{M} + (2N-2M) \frac{\mu}{M} \leq 1.$$
Taking $N = \sqrt{2} M$, we get 
$$(4-2 \sqrt{2}) \mu \leq 1 \ \mbox{so} \ \mu \leq 1/(4-2 \sqrt{2}).$$
<hr>
We can push further. Consider times $N_1$, $N_2=1.25992 N_1$ and $N_3= 1.5874 N_1$. It turns out $\mu$ is as large as possible if we subdivide intervals from time $N_1$ at all opportunities, rather than further subdividing already subdivided intervals. This means that, at time $N_3$, there will be $N_1-(N_3-N_1)=2 N_1-N_3$ intervals of size at least $\mu/N_1$, $2 N_2-2 N_1$ of size at least $\mu/N_2$ and $2 N_3-2N_2$ of size at least $\mu/N_3$. This gives $1.2378 \mu \leq 1$ or $\mu \leq 0.808$.