Finding Zin of non ideal non-inverting amplifier
It should be noted that \$Z_{in}\$ means the input impedance at IN+ (and not IN-).

Due to negative feedback, the IN- voltage will be pulled towards IN+. The voltage across \$R_{id}\$ is therefore kept very small, causing only a very small current to flow from IN+ to IN-. So because of feedback, the resistor \$R_{id}\$ looks bigger. We are dealing with series-shunt feedback!

Basic feedback theory will tell you that in this case \$R_{id}\$ needs to be multiplied by \$1 + \beta A\$ (one plus the loopgain). So what you see is a resistor to ground (\$2R_{icm}\$) in parallel with a resistor to IN- modified by feedback (\$R_{id}(1 + \beta A)\$).

To be correct, the equivalent resistance to ground at IN- still has to be added in series to the modified \$R_{id}(1+\beta A)\$, but this contribution will probably be much smaller.

So you end up with a total input impedance of approximately

\$Z_{in} \approx 2R_{icm}||\left[R_{id}(1 + \beta A(s))\right]\$

EDIT:


In order to get the same formulas as the book for \$V_x\$ (the voltage at IN-), you need to assume that \$R_{id} \gg R_{icm} \gg R_1,R_2\$ and \$R_o\approx 0\$. So this means that \$R_{id}||R_{icm}\approx R_{icm}\$. By blocking the path via \$R_{id}\$, \$V_i\$ cannot influence IN- directly.
\$I_x\$ is the current flowing from IN+ to IN-, and so is also equal to the total current seen flowing to "ground" at IN- as far as IN+ is concerned. So if we were to replace everything at IN- by a single impedance to ground (which is the \$Z_x\$ we're trying to calculate), then the current through that impedance would be the same as through \$R_{id}\$.
\$Z_x\$ is the total equivalent impedance to ground at IN-. This is the voltage \$V_x\$ divided by the current to ground \$I_x\$.
And so finally, you can find the total equivalent impedance to ground as \$R_{id}+Z_x\$ in parallel with the shown \$2R_{icm}\$.


As you can see, the formulas reflect again that the current is artificially made small by the feedback. \$V_x\$ is brought closer to \$V_i\$ by \$V_o\$, and this reduces the current through \$R_{id}\$ compared to an \$R_{id}\$ to ground.