The sphere $S^d$ is Euclidean space $E^d$ with infinity identified as a single point I'm reading about anti de Sitter spacetime, and I found the following statement:

$$ds^2 = \frac{1}{\cos^2 \psi} \big( -dt^2 + d\psi^2+ \sin^2 \psi d\Omega_{d-2}^2 \big).$$
Thus, the spatial sections of $AdS^d$ (constant $\psi$) are bounded by $S^{d-2}$ ($\psi \rightarrow \pi/2$), which may be thought of as Euclidean space $E^{d-2}$ with spatial infinity identified as a single point. Adding back the time coordinate, we conclude that the anti de Sitter spacetime $AdS^d$ is bounded by $M^{d-2,1}$.

[Zee, Einstein's gravity in a nutshell, ch. IX.11, page 655]. I don't see this. I understand that $S^{d-2}$ can be thought of as $E^{d-2}$ with infinity identified in a point (like in the stereographic projection), but this is topologically, not metrically. I mean, this doesn't mean that they have the same metric. Actually, for constant $\psi$ you get the metric
$$ds^2 = C_1 \big( -dt^2 + C_2 \, d\Omega_{d-2}^2 \big) $$
where $C_1 = \frac{1}{\cos^2 \psi}$ and $C_2 =\sin^2 \psi$ are constants. This is not a flat metric, but rather a $(d-2)$-sphere with time. So it can't represent Minkowski spacetime $M^{d-2,1}$. Why does he say so?
 A: Maybe the author was being lax about the distinction between flat and conformally flat — that is, about the distinction between Minkowski spacetime and its conformal compactification. This is plausible, because AdS does not have a "boundary" in the strict sense. The "boundary" of AdS spacetime is a really feature of the conformal compactification of AdS. According to page 2 in [1]:

the boundary of the conformal compactified AdS$_{d+1}$ is identical to the conformal compactification of the $d$-D Minkowski space, i.e., $\mathbb{R}\times S^{d-1}$. ... The $d$-D Minkowski space $\mathbb{R}^{1,d-1}$ is conformally compactified to $\mathbb{R}\times S^{d-1}$.

Considering the boundary metric only up to conformal equivalence is natural in this context, as emphasized on page 3 in [2]:

the metric on the boundary [of AdS] is only specified up to rescaling, i.e. a Weyl transformation.

A relatively intuitive account is given on page 4 in [3]:

Since null geodesics do not care about the Weyl factor of the metric, ie multiplication by scalar function, we see that the causal structure of AdS$_{d+1}$ will be the same as that of a "solid cylinder" with metric
  $$
  ds^2=-dt^2+d\rho^2+\sin^2\rho\,d\Omega_{d-1}^2.
$$
  ... Perhaps the most important feature of AdS space is its asymptotic
  boundary at $r = \infty$ (or at $\rho = \pi$), which has topology $\mathbb{R}\times S^{d-1}$. ... signals can be sent to this boundary and "replies" received in a finite proper time according to a massive observer sitting at rest in the center of the space. This suggests that despite the infinite spatial volume of AdS, we might want to think of it as being something like a finite box.

Given all of this, it seems plausible that the author quoted in the OP described the "boundary" of AdS in terms of flat Minkowski spacetime because the boundary is a feature of the conformal compactification of AdS, so only the conformal structure matters.

References:
[1] "A Very Introductory AdS/CFT, "http://theory.uchicago.edu/~ejm/course/JournalClub/Basic_AdS-CFT_JournalClub.pdf
[2] https://ocw.mit.edu/courses/physics/8-821-string-theory-fall-2008/lecture-notes/lecture12.pdf
[3] "TASI Lectures on the Emergence of the Bulk in AdS/CFT, "https://arxiv.org/abs/1802.01040
