I/O Resistance of common source MOSFET with source degeneration
In general source degeneration resistor "adds" a negative feedback to the circuit (current-series feedback). In this case, we sample the output current (\$I_D\$) and return a proportional voltage in series with the input (\$V_{GS} = V_G - I_D*R_S\$).
This type of a feedback increases \$Rin\$ and \$Rout\$. But notice that the MOSFET itself has a very large \$Rin =\infty\$, therefore \$Rin = R1||R2\$ remains unchanged.  

The voltage gain also drops to \$Av = -\frac{R_D}{R_S + 1/gm} = -\frac{R_D||R_L}{\frac{1}{gm} +R_S||R_3}  \$ 

This also improves linearity, because without \$R_S\$ voltage gain is \$gm*R_D\$ and as you should know \$gm\$ varies with drain current. 
Because \$gm\$ is a function of drain current (\$I_D\$), the voltage gain will vary with signal swing and the voltage gain also. But if we add external source resistance \$R_S\$ we notice that the \$R_S\$ does not change with the signal swing (\$I_D\$ swing)so, the overall voltage gain is stabilized and is more linear. 

For \$R_S &gt;&gt; 1/gm\rightarrow A_V\approx \frac{R_D}{R_S}\$

Now let us look at \$rout\$. If we are looking from the load perspective we can see two paths for a  AC current to flow: 



First through \$R_D\$ resistor. 

And the second one through MOSFET channel -->\$R_S\$ into GND.

As you can see now \$R_S\$ resistor is in series with the MOSFET channel. 

So, to find resistance seen from the drain terminal into the MOSFET we need to use a small-signal-model. 



\$r_x = \frac{V_X}{I_X}\$ and because \$V_G = 0V\$ we have: 

$$V_{GS} = -I_X*R_S $$ 

And from KVL we have 

$$V_X = I_{ro}*ro+I_X*R_S$$ 

$$I_{ro} = I_X - gm*V_{GS}$$ 

$$V_X=\left ( I_X - \left (gm\left ( -I_X \right )R_S  \right ) \right )ro + I_XR_S $$

And solve for \$I_X\$
$$I_X = \frac{V_X}{ R_S + ro + gm*R_S*ro} $$ 
And finally we have 
$$r_x = R_S + ro + gm*R_S*ro = ro(1+gmR_S+\frac{R_S}{ro}) $$

$$r_x =  ro*(1+gmR_S)+R_S $$

As you can see adding \$R_S\$ resistor increase the MOSFET resistance. 

The \$ro\$ is boosted by a factor of \$(1+gm R_S)\$ 

So, the overall \$r_{out}\$ is equal to: 

$$r_{out} = R_D||r_x $$ 

and because \$R_D&lt;&lt;r_x\$ we have \$r_{out} \approx R_D\$