Arriving at a wrong output impedance for a BJT Emitter Follower Configuration Circuit
To be able to see the difference try to analysis these two circuits.  

The first one is CE amplifier 



And for this circuit 

\$Z_{OUT} = \frac{V_X}{I_X} = R_C\$ 

because \$I_B = 0A\$

But for the emitter follower, we have a different situation:



And \$ I_B\$ is not equal to \$0A\$ despite the fact that the Vin = 0

For this circuit \$I_B = -\frac{V_X}{r_{\pi}}\$

And \$Z_{OUT} = \frac{V_X}{I_X}\$

Let us try to find the \$Z_{OUT}\$ for the equivalent circuit:

 

I hope that you see that \$R_E\$ is in parallel with the resistance seen from the emitter terminal into the BJT. 

And our test current is 

$$I_X = I_B + \beta I_B + I_{RE}$$ 

But if we ignore \$R_E\$ resistance for a moment we can find the transistor resistance seen from the emitter looking into BJT. 

$$I_X = I_B + \beta I_B = I_B(\beta +1)$$ 

Additional we know that \$I_B = \frac{V_X}{r_\pi}\$

we can write 

$$I_X = I_B(\beta +1)= \frac{V_X}{r_\pi}(\beta +1) = \frac{V_X (\beta +1)}{r_\pi} $$ 

$$\frac{I_X}{V_X} =\frac{\beta +1}{r_\pi} $$

And finally 

$$Z_{OUT} = \frac{V_X}{I_X} = \frac{r_\pi}{\beta +1}||R_E $$

and because 

$$r_\pi = (\beta +1)re $$

we have 

$$Z_{OUT} = re||R_E $$