How to calculate the output impedance of the BJT in a Common Collector Amplifier?
The trick with the CC circuit is that there's local feedback going on. 

Look at this schematic:



The output impedance is an indication of how much the voltage at \$V_{out}\$ would change if I draw a little bit of extra current out of that point.

What happens when I do that?

I'm drawing some extra current from the emitter. What would the emitter voltage do? It is not a hard voltage, it cannot supply unlimited current so the voltage at the emitter will decrease a little.

Now what happens?

That drop of the emitter voltage increases \$V_{BE}\$ a little. Increasing \$V_{BE}\$ means that the transistor is "opened further", it will supply more current from the collector.

This extra current from the collector is crucial as it flows to the emitter and tries to compensate for the extra current I was drawing there. This causes the CC circuit to have a very low output impedance. When designed properly this output impedance will be much lower than the value of \$R_L\$ (in your drawings it is \$R_E\$).

So you must take this effect of the collector current into account because it is significant.

As an experienced circuit designer I already know that the output impedance is roughly equal to \$\frac{1}{gm} = \frac{1}{40I_c}\$. At \$I_c\$=1mA that would give 25 ohms, which is generally much lower than the value of \$R_L\$

You might also want to look at this derivation which does not use the \$h\$ parameters.