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 >> 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<<r_x\$ we have \$r_{out} \approx R_D\$
