Why does the water dissociation constant Kw remains constant?
Your question reflects your knowledge in chemistry so I'll try to keep this explanation simple as possible.
Water conduct electricity because it contains $\ce{H+}$ and $\ce{OH-}$ ions by auto-ionization:
$$\ce{H2O &lt;=&gt; H+ + OH- \tag1}$$
It has been defined that this ionization is a constant at $\pu{25 ^\circ C}$. Thus by definition:
$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = 1.0 \times 10^{-14} \tag2$$
Accordingly, by definition for pure water, $ [\ce{H+}]=[\ce{OH-}]= 1.0 \times 10^{-7}$. And then, there is a widely accepted principle in chemistry called Le Chatelier’s principle:

Le Chatelier’s principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

Accordingly, increasing the concentration of reactants (species on left hand side of the reaction) will drive the reaction to the right (more products), while increasing the concentration of products (species on right hand side of the reaction) will drive the reaction to the left (more reactants). At constant temperature, even though concentrations of species (e.g., here they are $\ce{H+}$ and $\ce{OH-}$ in equation $(1)$) at new  equilibrium state is changed, the equilibrium constant remains same. In other word, the equilibrium constant is depends only on temperature.
According to Le Chatelier’s principle, in in equation $(1)$, if you add more $\ce{H+}$ or $\ce{OH-}$ (RHS), the equilibrium would be adjusted to reduce that amount by making more water. But since this addition is at constant temperature, $K_\mathrm{w}$ remains constant.