Probability that for two random points in unit ball, one is closer to the center than to the other point

$d(p_1,O)\lt d(p_1,p_2)$ or $d(p_2,O)\lt d(p_1,p_2)$ if and only if one of $d(p_1,O)$ and $d(p_2,O)$ is the least of the three distances. By symmetry, the probability for this is twice the probability that $d(p_2,O)$ is the least of the three distances.

Fix $p_1$ at $(0,r)$. Then $d(p_2,O)$ is the least of the three distances if $p_2$ lies within the circle of radius $r$ around the origin and below $y=\frac r2$, and thus in a circular segment with radius $r$ and angle $\frac{4\pi}3$, so the desired probability is

$$ 2\cdot2\cdot\frac1\pi\int_0^1r\mathrm dr\,\frac{r^2}2\left(\frac{4\pi}3+\frac{\sqrt3}2\right)=\frac23+\frac{\sqrt3}{4\pi}\approx80\%\;, $$

where one $2$ is the symmetry factor above, another $2$ normalises for $p_1$ and $\frac1\pi$ normalises for $p_2$.

P.S.: I just realised that I simply assumed that you meant the unit ball in $2$ dimensions but you hadn't actually specified the number of dimensions. In case you meant the unit ball in three dimensions, we need to use a spherical cap of height $\frac32r$ and adjust the radial density and the normalisation; the probability in this case is

$$ 2\cdot 3\cdot\frac3{4\pi}\int_0^1r^2\mathrm dr\,\frac{\pi\left(\frac32r\right)^2}3\left(3r-\frac32r\right)=\frac{81}{16}\int_0^1\mathrm dr\,r^5=\frac{27}{32}\approx84\%\;. $$

In case you wanted the result in arbitrary dimensions, you'll find some information about the volumes of the resulting hyperspherical caps with height $\frac32r$ at Stars in the universe - probability of mutual nearest neighbors.