placing balls inside ball

As pointed out by @mjqxxxx, this is possible.

In fact, it is possible to pack two families of open balls of radii $\frac12, \frac13, \ldots$ into unit sphere.
The picture below illustrate one possible configuration (with $n$ up to $300$):

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In above picture, the two families are related to each other by a reflection with respect to origin.

Let us concentrate on the families of brown balls. The brown ball with radius $\frac12$ is centered at $(0,0,-\frac12)$. For $n \ge 3$, the brown ball with radius $\frac1n$ is centered at

$$(x_n,y_n,z_n) = (r_n\sin\theta_n\cos\phi_n,r_n\sin\theta_n\sin\phi_n, r_n\cos\theta_n)$$

where $\displaystyle\;r_n = 1 - \frac1n,\;\theta_n = \frac{\pi}{4}\left(\frac{n+1}{n-1}\right)\;$ and $\phi_n$ is defined recursively by

$$ \phi_n = \begin{cases} 0, & n = 3\\ \phi_{n-1} + \rho_n + \rho_{n-1}, &n > 3 \end{cases} \quad\text{ where }\quad \rho_n = \sin^{-1}\left(\frac{1}{(n-1)\sin\theta_n}\right) $$ If one orthogonal project the ball with radius $\frac1n$ onto $xy$-plane, $\rho_n$ will be the half-angle subtended by the ball's image with result to origin. The condition $\phi_{n} - \phi_{n-1} = \rho_{n} + \rho_{n-1}$ guarantee the balls with radii $\frac{1}{n}$ and $\frac{1}{n-1}$ are disjoint from each other.

• For $n = 2$, the brown ball with radius $\frac12$ is touching the blue ball with radius $\frac12$ and the brown ball with radius $\frac13$. However, their interiors are disjoint.

• For small $n$, one can build a 3d model of above configuration and verify by eye the balls are disjoint from each other. Above picture contains balls with $n$ upto $300$, balls with $2 < n \le 300$ have been inspected and they are disjoint from each other.

• For larger $n$, the balls form a spiral converging to some sort of limit cycle at $\theta = \frac{\pi}{4}$. For a typical ball there, its projection onto $xy$-plane will subtend an angle $2\rho_n \sim \frac{2\sqrt{2}}{n}$ with respect to the origin. Furthermore, the balls nearby will be arranged info "arms". Let $\frac{1}{Kn}$ be the radius of nearest ball on the next arm of spiral (i.e. the arm closer to the limit cycle immediately above the current arm). The "angles" between the ball and its nearest neighbor on next arm will be around $2\pi$. This means $$\int_{n}^{Kn} \frac{2\sqrt{2}}{n} dn = 2\sqrt{2}\log K \approx 2\pi \quad\implies\quad K \approx e^{\pi/\sqrt{2}}$$

  The distance between this pair of balls is around $\theta_n - \theta_{Kn} \approx \frac{\pi}{2Kn} - \frac{\pi}{2n} = \frac{\pi}{2n}\left(\frac1K-1\right)$.
  We can compare it with the required separation $\frac{1}{n} + \frac{1}{Kn} = \frac{1}{n}\left(1 + \frac{1}{K}\right)$ between them.
  Since the ratio $\frac{\pi}{2}\left(\frac{K-1}{K+1}\right) = \frac{\pi}{2}\left(\frac{e^{\pi/\sqrt{2}}-1}{e^{\pi/\sqrt{2}}+1}\right) \approx 1.263418 > 1$, when $n$ is large, the ball with radius $\frac1n$ is disjoint from the one on next arm of spiral.

To bridge the analysis for small and large $n$, we need to check by the time $n$ reaches $300$, whether we can switch to use the result for large $n$ or not.

I don't have a rigorous proof. However, when we increase $n$ to around $100$, the relative error between $2\rho_n$ and its approximation $\frac{2\sqrt{2}}{n}$ already falls below $0.55\%$. The relative error in $K$ should be around $\frac{\pi}{\sqrt{2}} \cdot 0.55\% \sim 1.3\%$. As far as I can see, other approximations should generator relative errors of same order. Since the ratio $1.263418$ above is tens of percents larger than $1$. It seems $n = 100$ is already large enough.

Combining all these analysis, a safe bet is all the balls in above two families are disjoint from each other.