This is not at all a complete answer but a remark which can be improved. It is a completely rewritten and replaces bullshit (explaining the first comments.)
Suppose the answer to Segerman's question is no. There exists thus a counterexample given by a decreasing sequence $r_1\geq \dots \geq r_n$ of radii such that $\sum_{i=1}^n r_i^2=1/2$ and one can not fit $n$ circles with radii $r_1,\dots,r_n$ into a circle $C_1$ of radius $1$. Suppose $n$ is the smallest integer for which a counterexample exists. Then $r_1<2/3$. Indeed, the area $\pi(1/2-r_1^2)$ of the discs of radii $r_2,\dots,r_n$ is at most half the area $\pi(1-\rho_1)^2$ of the largest disc $C'$ which fits together with the disc $C_1$ of radius $r_1$ into $C$. Since $n$ is minimal, the $(n-1)$ circles of radii $r_2,\dots,r_n$ can be packed into $C'$.
This kind of argument can be improved (pack the circles of radii $r_2,\dots$ into more than one circle of suitable radii which fit into $C$ together with $ C_1$) in order to lower the upper bound on the largest radius of a minimal counterexample.
I guess one can use a packing argument showing that a solution always exists if the largest radius is small enough.
These two bounds can perhaps be made to met (but I fear that the involved combinatorics are quite messy).
Let me make the argument for small radii a little bit more (but not entirely, I agree that some more work is needed) rigorous. (This is probably similar to the argument suggested by fedja (see comment below), I admit that I do not quite understand the details).
Supposing the total area of all discs of radius $\leq \epsilon$ exceeds $2\pi \epsilon$, look at the annulus of width $\epsilon$ and outer radius $1$ inside the large disc, supposed to be of radius $1$. If $\epsilon$ is very small, such an annulus looks locally like a strip delimited by two parallel lines at distance $\epsilon$. I have thus to prove that given a collection of radii $\leq 1/2$, I can cover more than half of the area of a very long strip delimited by two parallel lines at distance $1$.
If all discs have equal radius, then I can cover asymptotically a proportion of at least $\pi/4\sim .7854$ of the strip (the least favorable case corresponds to a collection of discs all of maximal radius $1/2$). The general case should be better but is harder to analyze. I claim however that this analysis can be made. Indeed, up to subdividing the strip into smaller substrips, we can assume that all radii are $>1/4$ and we have then a fairly small number of combinatorial situations to consider. We can thus compute the worst case.