Does the number of H+ ions in solution go up on dilution of a weak acid?
Let's say we have got a solution of an acid with a pK$_a=5$. The resulting pH dependent on the concentration of that acid would be like:

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Examples: $0.1~\ce{M}$: pH$=3.0$; $0.001~\ce{M}$: pH$=4.0$; $0.00001~\ce{M}$: pH$=5.2$; $0.0000001~\ce{M}$: pH$=6.8$

Depending on the pH resulting from these example concentrations we can have a look at percentage composition of the whole acid base system. (Degree of association $\alpha'$ in blue, degree of dissociation $\alpha$ in light orange.)

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Examples: pH$=3.0$: $\alpha\approx 1.0~\%$; pH$=4.0$: $\alpha\approx 9.5~\%$; pH$=5.2$: $\alpha\approx 61.8~\%$; pH$=6.8$: $\alpha\approx 98.4~\%$

One can see that low acid concentrations lead to higher pH values than higher acid concentrations which automatically result in an increasing degree of dissociation.

So for me this is no big intuition thing but rather a small sequence of actions.



Lets say it in math:

The proton concentration for your system can be described by $$c(\ce{H3O+}) = \sqrt{c_0(\ce{HA})~k_a+k_W}$$ (adding $k_W$ for very low concentrations) and the equation for the degree of dissociation is $$\alpha = \frac{k_a}{k_a+c(\ce{H3O+})}$$ which can be combined to give $$\alpha\left(c_0(\ce{HA})\right)=\frac{k_a}{k_a+\sqrt{c_0(\ce{HA})~k_a+k_W}}$$



One could expand this system a little more (using the exact pH equation including the self-ionization of water) and come up with a thorough scheme like the following.

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