Trouble understanding output impedance of common collector
KCL Analysis, sans Intuition
Let's start by ignoring intuition for a moment and just working through the problem. To begin, the schematic:



simulate this circuit &ndash; Schematic created using CircuitLab
(For those interested, I've provided the OP's fuller context at the end, below.)
Normally, for small signal purposes, you might also insert \$r_e\$ in the above circuit right at the tip of the emitter of \$Q_1\$. But the textbook is ignoring its value until section 2.3 and considering it absent for now.
You know that \$V_\text{B}-V_\text{E}=V_\text{BE}\$ and that for small signal purposes, absent \$r_e\$, this is a fixed voltage difference. This allows us to substitute one for the other.  Also, note that \$I_\text{E}=\frac{V_\text{E}}{R_\text{E}}\$.
Assuming \$I_x\$ is an arbitrary current sink load that will be either \$0\:\text{A}\$ (no load) or \$1\:\text{A}\$ (loaded), then the KCL is:
$$\begin{align*}
\frac{V_\text{E}+V_\text{BE}}{R_1}+\frac{V_\text{E}+V_\text{BE}}{R_2}+\frac{V_\text{E}+V_\text{BE}}{Z_\text{IN}}+\frac{\frac{V_\text{E}}{R_\text{E}}+I_x}{\beta+1}&amp;=\frac{V_\text{CC}}{R_1}+\frac{0\:\text{V}}{R_2}+\frac{V_\text{IN}}{Z_\text{IN}}+\frac{\frac{0\:\text{V}}{R_\text{E}}}{\beta+1}
\end{align*}$$
(In the above, I've placed the outflowing currents on the left side and the inflowing currents on the right side. Although I've written about this novel approach to KCL many times, a more recent example is shown here.)
So the above equation says

&quot;The current flowing out of the base node through \$R_1\$, plus the current flowing out of the base node through \$R_2\$, plus the
current flowing out of the base node through \$Z_\text{IN}\$, plus
the current flowing out of the emitter node through \$R_\text{E}\$
as seen by the base node is equal to the current flowing into the base node from \$V_\text{CC}\$ through \$R_1\$, plus the current
flowing into the base node from ground through \$R_2\$, plus the
current flowing into the base node from \$V_\text{IN}\$ through
\$Z_\text{IN}\$, plus the current flowing into the emitter node from
ground through \$R_\text{E}\$ as seen by the base node.&quot;

If you solve the above for \$V_{\text{E}\left(I_x\right)}\$, then you can work out: \$Z_\text{OUT}=\frac{\Delta \,V_\text{E}}{\Delta\,I_\text{E}}=\frac{V_{\text{E}\left(I_x=0\right)}-V_{\text{E}\left(I_x=1\right)}}{1\:\text{A}-0\:\text{A}}\$:
$$Z_\text{OUT}= \frac{\frac1{\beta+1}\left(\beta+1\right)R_\text{E}\,R_1\, R_2\,Z_\text{IN}}{\left(\beta+1\right)R_\text{E}\,R_1\, R_2+\left(\beta+1\right)R_\text{E}\,R_1\,Z_\text{IN}+\left(\beta+1\right)R_\text{E}\, R_2\,Z_\text{IN}+R_1\, R_2\,Z_\text{IN}}$$
That's exactly the same result you'd get if you took \$\frac1{\beta+1}\left[R_1\mid\mid R_2\mid\mid Z_\text{IN}\mid\mid \left(\beta+1\right)R_\text{E} \right]\$ or, multiplying \$\frac1{\beta+1}\$ through:
$$Z_\text{OUT}=\left[\left(\frac{R_1\mid\mid R_2\mid\mid Z_\text{IN}}{\beta+1}\right)\mid\mid R_\text{E} \right]$$
The only difference here from the textbook being that the authors chose to use \$\beta\$ as an approximation for \$\beta+1\$.
Intuition
Look back at the original schematic. There, you can readily see that \$R_1\$, \$R_2\$, and \$Z_\text{IN}\$ are all tied from a voltage source (assumed ideal) to a shared node at the BJT base. From the point of view of the base, looking at those three impedances from an AC standpoint, they are all in-effect &quot;grounded&quot; and therefore &quot;in parallel&quot; with each other.
Now, since tiny current variations at the base imply much larger current variations at the emitter, the parallel resistance seen at the base will look \$\beta+1\$ times smaller at the emitter. This is then taken in parallel to \$R_\text{E}\$.
So that's where an intuitive view comes from.
The Art of Electronics, 3rd edition, Page 84
The worked problem you cite has \$V_\text{CC}=+15\:\text{V}\$, \$R_1=130\:\text{k}\Omega\$, \$R_2=150\:\text{k}\Omega\$, \$Z_\text{IN}=10\:\text{k}\Omega\$, \$R_\text{E}=7.5\:\text{k}\Omega\$ and \$\beta=100\$. With those values, you should find that \$Z_\text{OUT}\approx 85.59\:\Omega\$ and \$A_v\approx 0.86446\$. The book writes that \$Z_\text{OUT}\approx 87\:\Omega\$, which is quite close enough.
As the book also points out, since the design is for \$I_\text{E}\approx 1\:\text{mA}\$, then the dynamic Ebers-Moll AC impedance value they will discuss later (\$r_e\$) will be about \$26\:\Omega\$. (They say \$r_e\approx 25\:\Omega\$.) This is added, in series and will increase \$Z_\text{OUT}\$ to \$Z_\text{OUT}\approx 112\:\Omega\$. (The book writes it as \$110\:\Omega\$ using their slightly smaller value.)
Getting overly precise is pointless, so the textbook is handling this just as you should: showing at most two digits of precision.
Fuller Context taken from The Art of Electronics, 3rd edition:
The OP failed to provide the worked example that was at question, I think:

They are relying upon a simpler BJT model that does NOT yet include \$g_m\$ and upon earlier discussions about \$Z_\text{IN}\$ and \$Z_\text{OUT}\$ that also help frame the above discussion in the textbook.