How to prove at what pH the color of an indicator will change based on its pKa?
Thank you guys for your answers! I am fairly confident I have found the correct answer, or at least one of the possible correct answers. Here's it goes:

The indicator reacts with water according to the following reaction equation:
$$\ce{HA + H2O &lt;=&gt; H3O+ + A-}$$
Then using the Henderson-Hasselbach equation, I arrive at the following point:
 $$ K_s = \frac{[\ce{H3O+}]*[\ce{A-}]}{[\ce{HA}]} $$
Now, at the equivalence point, the concentration of the indicator an it's conjugate base are equal, hence:
 $$\require{cancel} K_s = \frac{[\ce{H3O+}]*\cancel{[\ce{A-}]}}{\cancel{[\ce{HA}]}} $$
$K_s = [\ce{H3O+}]$. After taking the negative log of both sides I arrive at the equation: $pK_s = pH$ (which was given to be 6).

Now, when the solution is acidic(pH lower than 6 and high [$\ce{H3O+}$] concentration), Le Chatelier sais then the balance will shift to the left of the top most equation. Favoring the indicator $\ce{HA}$ as opposed to it's conjugate base. Naturally, if the solution contains more indicator molecules, then it will also take on the indicators color. 


 Analogue to the previous reasoning, when the pH reaches values above those of 6, Le Chatleier sais that the reaction will favor the formation of the indicators conjugate base [$\ce{A-}$](the right side) and the solution takes on the color accordingly.