Percent Dissociated from Titration Curve
I'm not sure I follow your logic.
For a monobasic acid S $(\ce{HA})$ dissociation degree $α$ is

$$α = \frac{[\ce{H+}]}{c_\mathrm{a}},$$

where $c_\mathrm{a}$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_\mathrm{b}$:

$$c_\mathrm{a}V_\mathrm{a} = c_\mathrm{b}V_\mathrm{b} \implies c_\mathrm{a} = \frac{c_\mathrm{b}V_\mathrm{b}}{V_\mathrm{a}}$$

Finally, taking $V_\mathrm{a} = V_\mathrm{b} = \pu{50 mL}$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[\ce{H+}] \approx \pu{1e-4 mol L-1}$:

$$α = \frac{[\ce{H+}]V_\mathrm{a}}{c_\mathrm{b}V_\mathrm{b}} = \frac{\pu{1e-4 mol L-1}\cdot\pu{50 mL}}{\pu{0.1 mol L-1}\cdot\pu{50 mL}} = \pu{1e-3}~\text{or}~0.1\%$$