Question regarding Kleppner Kolenkow Example 5.5 "Escape Velocity—the General Case"

In example 5.5 from Kleppner Kolenkow (2nd edition), the general case of finding the escape velocity of a mass $m$ projected from the earth at an angle $alpha$ with the vertical, neglecting air resistance and Earth’s rotation, is presented.

Their analysis goes as follows:

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The force on $m$, neglecting air resistance, is

$$mathbf{F} = -mgdfrac{R_e^2}{r^2}mathbf{hat{r}}$$

where $mathbf{hat{r}}$ is a unit vector directed radially outward from earth’s center, $R_e$ is the radius of the Earth and $r$ is the distance of $m$ from the center of the Earth.
We don’t know the trajectory of the particle without solving the problem in detail, but for any element of the path the displacement $dmathbf{r}$ can be written as
$$dmathbf{r} = dr mathbf{hat{r}} + rdtheta boldsymbol{hat{theta}}$$
where $boldsymbol{hat{theta}}$ is a unit vector perpendicular to $mathbf{hat{r}}$, see the picture below for a sketch of the images they present.

Because $mathbf{hat{r}} cdot boldsymbol{hat{theta}} = 0$ we have
$$mathbf{F} cdot dmathbf{r} = -mg dfrac{R_e^2}{r^2}mathbf{hat{r}} cdot (dr mathbf{hat{r}} + rdtheta boldsymbol{hat{theta}}) =
-mgdfrac{R_e^2}{r^2}dr.$$
The work-energy theorem becomes
$$dfrac{m}{2}(v^2-v_0^2) = -mgR_e^2 int_{R_e}^r dfrac{dr}{r^2} =
-mg R_e^2(dfrac{1}{r} – dfrac{1}{R_e})$$
Here is where I get lost: They say the escape velocity is the minimum value of $v_0$ for which $v=0$ when $r to infty$. We find
$$v_{text{escape}} = sqrt{2gR_e} = 1.1 times 10^4 text{m/s}$$
which is the same result as in the example when $alpha=0$, that is when the mass is projected straight up (presented earlier in the book). They write, "In the absence of air friction, the escape velocity is independent of the launch direction, a result that may not be intuitively obvious".

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Indeed I find this very unintuitive and have one gripe with the analysis presented. I don’t understand how we can assume that $r$ even will go to infinity in the first place. As far as I can see, what have been shown is that if $r to infty$, then $v_0 = sqrt{2gR_e}$ will make $v=0$. What I don’t see have been shown, however, is that setting $v_0 = sqrt{2gR_e}$ will make $r to infty$, which I think really is what escape velocity should be. (This is what was done in the case of a pure vertical projection of the mass, it was shown that if $v_0 = sqrt{2gR_e}$ then $r_{text{max}} to infty$ but this has not been done in the more general case presented here I think.)

Question
Based on this similar question the curvature of the Earth seems to make it so that even if you fire the mass horizontally it won’t crash down or simply stay at a constant height above the earth, but this would have to be shown in the analysis right, it is not merely enough to assume that $r$ can go to infinity?