The claim is not that the (surface) area, $4 \pi r^2$, of a sphere of radius $r$ grows more slowly than the area, $\pi r^2$, of a disk of radius $r$ on the plane. Rather, the book is asserting that the area of a disk on a sphere grows more slowly with radius than does a disk on the plane.
If we pick a point $P$ on a sphere of radius $R$, we can ask which points on $R$ are a fixed distance $r$ from $P$ along the surface of the sphere---(for $0 < r < \pi R$) by symmetry these points comprise a circle. The area the book means is the region inside this circle (i.e., in the region bounded by the circle and containing $P$).
Now (for $0 \leq r \leq \pi R$), the central angle between $P$ and a point a distance $r$ from $P$ along the sphere is just $\frac{r}{R}$. Thus, if we declare $P$ to be the north pole of the sphere and use the usual colatitude-longitude coordinates on the sphere, we find that the area $A_{\text{sphere}}(r)$ of the region $D$ of points with a distance $r$ from $P$ is
$$A_{\text{sphere}}(r) = \iint_D dA = \int_0^{2 \pi} \int_0^{\frac{r}{R}} \sin \phi \,d\phi \,d\theta = 2 \pi R^2 \left(1 - \cos \frac{r}{R}\right).$$
Now, the rate of growth (w.r.t. radius $r$) of the area $A_{\text{plane}}(r)$ of a disk on the plane is
$$\frac{d}{dr} A_{\text{plane}} (r) = 2 \pi r,$$ and the rate of growth of the region $D$ with respect to the radius $r$ measured along the surface of the sphere is
$$\frac{d}{dr} A_{\text{sphere}} (r) = 2 \pi R \sin \frac{r}{R}.$$
Since $0 < \sin u < u$, for $u > 0$, we have for $0 < r < \pi R$ that
$$\frac{d}{dr} A_{\text{sphere}} (r) = 2 \pi R \sin \frac{r}{R} < 2 \pi R \left(\frac{r}{R}\right) = 2 \pi r = \frac{d}{dr} A_{\text{plane}} (r)$$
as claimed.
Note that no point is further than $\pi R$ from $P$ (and only the point $-P$ antipodal to $P$ is exactly that distance away). So, the claim is trivially true for $r > \pi R$: By that point, the sphere is already covered but the disk in the plane keeps growing.
Remark We can extract a little more information by inspecting the Taylor series of the ratio $$\rho(r) := \frac{A_{\text{sphere}}(r)}{A_{\text{plane}}(r)}$$ around $r = 0$. Expanding to third order gives
$$\rho(r) = 1 - \frac{1}{12 R^2} r^2 + O(r^4).$$ Now, we can read off the Gaussian curvature of the sphere at the point $P$ by inspecting the quadratic term; importantly, we can do this for any surface, so comparing the rates of growth of disks on general surfaces with those on the plane is important insofar as it yields a reasonably concrete interpretation of curvature.
Here the $4\pi$ should be viewed not as $4$ times $\pi$ but rather as $2$ times $2\pi$ and then there is a reasonably nice geometric explanation. Namely, if you project the sphere to, say, the $z$-axis, then the inverse image of each small subsegment of $[-1,1]$ will have area $2\pi h$ where $h$ is the length of the subsegment. You see this particularly clearly in the case of a small subsegment around $0$ in which case you get a thin strip which is roughly the product of the equator (of length $2\pi$) by a segment of length $h$. Therefore the total area is $2\times 2\pi=4\pi$.
Four unit disks do have the same area as the unit sphere, but you can't patch them together without distorting them so as to form the sphere. One way of seeing it is in terms of a metric invariant called Gaussian curvature (this invariant vanishes for the disk but is equal to 1 for the sphere).
Best Answer
No. It just means that (hyper)cubic lattice sphere packing (where the centers of the spheres are placed in a cubic grid, say spheres with radius $\frac12$ centered at each point with integer cartesian coordinates) is very inefficient in higher dimensions, and the room between the spheres become large enough to fit even larger spheres.
Counterintuitive? Yes, but mostly because we are relatively low-dimensional beings with limited imagination. Paradox or contradiction? No.