Degree of dissociation of weak electrolyte at infinite dilution
There are a number of ways to explain this. In words, the entropic gain with dissociation overwhelms any attraction between counterions as the electrolyte is diluted. 

Mathematically: for a m:n electrolyte undergoing dissociation according to the reaction

$$\ce{A_mB_n -&gt; mA^{\nu_A -} + nB^{\nu_B +}}$$

if the initial concentration of electrolyte $\ce{A_mB_n}$ is $c_{T}$, define the degree of dissociation as

$$\alpha = \frac{c_{AB}}{c_{T}}$$

where $c_{AB}$ is the concentration of dissociated $\ce{A_mB_n}$. Assuming ideality and unit molarity as standard concentration, the equation for the equilibrium constant can be written as $$K=\frac{c_A^m c_B^n}{c_{AB}}$$ and solved in terms of the degree of dissociation. For a 1:1 electrolyte this leads to 

$$\alpha = \frac{\beta}{2}\left[\left(1 + \frac{4}{\beta} \right)^{1/2} -1\right] \tag{1}$$ 

where $\beta = K/c_T$. As $c_T \rightarrow 0$ it follows from this last expression that $\alpha  \rightarrow 1$ #QED

PS The following equation (from Wikipedia's page on Ostwald's law of dilution) $$\alpha=\sqrt{\frac{K}{c_T}} \tag{2}$$ does not hold over the entire concentration range $c_T$. It can be obtained from Eq. (1) by assuming $K&lt;&lt;1$ and therefore $\beta&lt;&lt;1$. For increasingly dilute solution Eq. (2) fails and  $\alpha  \rightarrow 1$.