Sequential addition of points on a circle, optimizing asymptotic packing radius
This is really a comment, but it's getting a bit long for the comment box.
I want to point out that in addition to what you call the "greedy process," there's another obvious attempt, which could be called the "stingy process."
Let a target $\mu$ be given. There's basically one rule, namely don't squander unused space unless absolutely forced to. More formally, if I have already chosen points $0\le x_1, \ldots , x_N\le a$, then these define $N$ intervals $I_j$. I must now place the next point such that the minimum distance is $\ge\mu/(N+1)$ (I will keep the whole sequence $\ge\mu$, not just the $\liminf$). If I have an interval of length at least twice this distance, then I put my point inside such an interval (let's say inside the smallest one that works and I will also make one distance equal to $\mu/(N+1)$, though that might not be optimal). If not, then I grudgingly add an interval of that length to the right end of the current configuration.
This gives me intervals that get subdivided, and every once in a while a new interval gets added at the current right endpoint. The whole procedure will be a success if the limit of these right endpoints is $\le 1$.
I fooled around some with this; let's take $\mu=1$. Then the first few points are $$ 0, 1/2, 1/2+1/3, 1/4, 1/2+1/3+1/5, 1/2+1/6, \ldots $$ It seems that $\lim a_N =1/2+1/3+1/5+1/7+1/11+1/14+\ldots \simeq 1.4$. If this is correct (but I'm not making any strong claims it is), then it means that we can get $\mu\simeq 1/1.4\simeq 0.7$, which is better than the golden mean.
In any event, it should be easy (for someone more computer savvy) to do the numerics and get a more reliable estimate.
I will show $\mu = \tfrac{1}{\log 4}$.
I first prove the upper bound $\mu \leq \tfrac{1}{\log 4}$. Fix a positive integer $r$. For $0 \leq k \leq r-1$, let $N_k$ be $2^{k/r} N$ rounded to the nearest integer, where $N$ is a large parameter. Considering the $N_{r-1}$ intervals present at time $N_{r-1}$, let $X_k$ be the number of them created between times $N_{k-1}$ and $N_k$; with $X_0$ the number created before time $N_0$. Then $$X_0 \frac{\mu}{N_0} + X_1 \frac{\mu}{N_1} + \cdots + X_{r-1} \frac{\mu}{N_{r-1}} \leq 1.$$ Set $S_k=X_0+X_1+\cdots +X_k$; we can rewrite the above equation as $$\mu \left( S_0 \left(\frac{1}{N_0} - \frac{1}{N_1}\right) + S_1 \left(\frac{1}{N_1} - \frac{1}{N_2} \right) + \cdots + S_{r-2} \left(\frac{1}{N_{r-2}} - \frac{1}{N_{r-1}}\right) + S_{r-1} \frac{1}{N_{r-1}} \right) \leq 1.$$
Now, $S_k$ is the number of those intervals present at final time $N_{r-1}$ which were created before time $N_k$, so we have $S_k \geq N_k - (N_{r-1} - N_k)=2 N_k - N_{r-1}$. The coefficient $\tfrac{1}{N_k} - \tfrac{1}{N_{k+1}}$ on the left hand side is positive, so we have $$\mu \left( (2N_0-N_{r-1}) \left(\frac{1}{N_0} - \frac{1}{N_1}\right)+ (2N_1-N_{r-1}) \left(\frac{1}{N_1} - \frac{1}{N_2}\right)+\cdots+ (2N_{r-2}-N_{r-1}) \left(\frac{1}{N_{r-2}} - \frac{1}{N_{r-1}}\right)+\frac{N_{r-1}}{N_{r-1}} \right) \leq 1.$$
We gather terms with the same denominator to obtain $$\mu \left( \frac{2N_0-N_{r-1}}{N_0} + \frac{2 N_1 - 2 N_0}{N_1} + \cdots + \frac{2 N_{r-1} - 2 N_{r-2}}{N_{r-1}} \right) \leq 1$$ or $$\mu \left( 2r - \frac{N_{r-1}}{N_0} - \frac{2 N_0}{N_1} - \cdots - \frac{2 N_{r-2}}{N_{r-1}} \right) \leq 1.$$ Plugging in our optimized values: $$\mu \left( 2r - r 2^{(r-1)/r} \right) \leq 1$$ so $$\mu \leq \frac{1}{2r (1-2^{-1/r})}.$$
Finally, taking the limit as $r \to \infty$ gives $\mu \leq \tfrac{1}{\log 4}$ as promised.
Now, for the lower bound. Again, fix a positive integer $r$. Divide the circle into $r$ equal arcs; each arc will be subdivided into intervals. At any point in our process there will be some integer $k$ such that one arc is divided into some intervals of length $2^{-k/r}$ and some of length $2^{-(k+r)/r}$; the other $r-1$ arcs will contain intervals of only one length, namely $2^{-j/r}$ for $k<j<k+r$. (Obviously, I am ignoring rounding issues.) As time ticks on, we subdivide the intervals in the first arc until they are all split in half, then move to the next arc, and so on.
This achieves (again, ignoring rounding) $$\mu = 2^{-(k+r)/r} \left( \frac{2^{k/r}}{r} +\frac{2^{(k+1)/r}}{r} + \cdots \frac{2^{(k+r-1)/r}}{r} \right) = \frac{1}{2r(2^{1/r}-1)}.$$ Again, sending $r \to \infty$ achieves $\tfrac{1}{\log 4}$.
Here is a slicker, though less motivated way, to achieve the lower bound. We will number the points starting with $x_0$ at position $0$. For $n >0$, write $n = 2^q+r$ with $0 \leq r <2^q$ and put $x_n$ at $\log_2\left(\tfrac{2n+1}{2^{q+1}}\right)$. So the first few points are $0$, $\log_2(1+1/2)$, $\log_2(1+1/4)$, $\log_2(1+3/4)$, $\log_2(1+1/8)$, $\log_2(1+3/8)$, $\log_2(1+5/8)$, $\log_2(1+7/8)$, ... When point $x_n$ is inserted, the smallest interval will either be the one just to the right of $x_n$ or else the furthest right interval. Approximating $\log$ by its linearization, these have lengths roughly $\tfrac{2^{q+1}}{(2n+1) \log 2} \tfrac{1}{2^{q+1}}=\tfrac{1}{(2n+1) \log 2}$ and $\tfrac{1}{2 \log 2} \tfrac{1}{2^q}=\tfrac{1}{2^{q+1} \log 2}$ respectively.
Multiplying these by $n+1$ (the number of intervals) gives $\tfrac{n+1}{(2n+1) \log 2}$ and $\tfrac{n+1}{2^{q+1} \log 2}$. The former approaches $\tfrac{1}{\log 4}$ while the latter oscillates between $\tfrac{1}{\log 2}$ and $\tfrac{1}{\log 4}$.
In this second answer, I want to discuss an upper bound on $\mu$ (not optimal, and I don't think this argument could give an optimal bound, even after fine tuning).
Suppose we have placed $N$ points such that (as required) the corresponding $N$ intervals $I_j$ have lengths $|I_j|\ge \mu/N$. Let's now position the next $N$ points. More specifically, let's denote by $k_j\ge 0$ the number of new points that I put into $I_j$. For this to be possible, I obviously need $|I_j|\ge (k_j+1)\mu/(2N)$.
Let $1\le S\le N$ be the number $j$'s with $k_j\ge 1$ (that is, the number of intervals that I actually use). Then $$ (N-S)\frac{\mu}{N} + \sum (k_j+1)\frac{\mu}{2N} \le 1 , $$ and since $\sum k_j=N$, this gives that $$ \mu\le \frac{2}{3-S/N} .\quad\quad\quad\quad (1) $$ Of course, this is not useful if $S/N\simeq 1$.
So we now need to look at this case $S\ge (1-a)N$ separately. I want to bound from below the lengths of these $S$ intervals. The worst case scenario arises when we first position the $aN$ remaining points, and then put $S$ points into that many new intervals. Then the $j$th such interval must have length $\ge 2\mu/(N+aN+j)$, for a total length of at least $$ 2\mu \sum_{j=1}^{(1-a)N} \frac{1}{(1+a)N+j}\simeq 2\mu \int_0^{(1-a)N} \frac{dx}{(1+a)N+x} =2\mu\log \frac{2}{1+a} . $$ Thus $\mu\le \frac{1}{a+2\log 2/(1+a)}$ (the extra $a$ comes from the length $\ge a\mu$ of the remaining $Na$ intervals), and when combined with (1), this gives the unconditional bound $$ \mu\le\max \left\{ \frac{2}{2+a} , \frac{1}{a+2\log 2/(1+a)} \right\} , $$ valid for all $0<a\le 1$.
The optimal $a$ seems to be around $a=0.3$, and then this gives $\mu\le 0.87$.