Why does output impedance behaves this way
When we talk about the small signal equivalent output impedance, we are talking about the differential impedance. How much does the voltage have to change to produce a certain change in the output current. 

Any resistor divider will have essentially the behavior you describe:



The equivalent output resistance of the divider is \$\dfrac{\mathrm{d}V_o}{\mathrm{d}I_L}\$. The KCL equation for the node Vo is
$$\frac{V_o-V_{cc}}{R_c} + \frac{V_o}{R_o} + I_L = 0$$
Rearranging terms
$$V_o = \left(\frac{1}{R_c}+\frac{1}{R_o}\right)^{-1}\left(\frac{V_{cc}}{R_c}-I_L\right)$$

The term with \$V_{cc}\$ doesn't affect the derivative so we have
$$R_{eq} = \dfrac{\mathrm{d}V_o}{\mathrm{d}I_L} = \left(\frac{1}{R_c}+\frac{1}{R_o}\right)^{-1}I_L$$ 

where I've fudged the sign, because I should have drawn the \$I_L\$ source facing the other way to begin with.

And that's just the formula for resistors in parallel so 
$$R_{eq} = R_c || R_o$$

Another way we remember this is that the voltage source (V1) has an equivalent resistance of 0 in the small-signal circuit, and after replacing V1 with a short circuit to form the small signal circuit we immediately see that \$R_c\$ and \$R_o\$ are in parallel.