equilibrium - Does the number of H+ ions in solution go up on dilution of a weak acid?

In my textbook, a footnote says:

In case of weak acids, on dilution the total number of $\ce{H^{+}}$ ions in solution increases because dissociation of the weak acid increases

This didn't make sense to me however. Consider the general dissociation reaction of a weak acid:

$$\ce{HA (aq) + H_2O \rightleftharpoons H_3O^{+} (aq) + A^{-}(aq)}$$

Now, the Chatelier's principle tells us that if we add more reactants ($\ce{H_2O}$), the reaction will proceed more to the right, and so the dissociation of the acid will increase, but the part about the total number of $\ce{H^{+}}$ ions in water being higher was still unclear to me.

Delving into the math a bit, we can see that

$$K_a = \frac{[\ce{H_3O^{+}}][\ce{A^{-}}]}{[\ce{HA}]}$$

Since the concentrations of the two products are equal,

$$[\ce{H_3O^{+}}] = \sqrt{[\ce{HA}]\cdot K_a}$$

Now, let us assume that we have a $1 \ M$ solution of the acid in question dissolved in 1 litre of water, plugging this into the equation, we have a $\sqrt{K_a}$ molar solution of $H^{+}$ ions. Since only 1 litre of solvent is there, we have $\sqrt{K_a}$ moles of ions.

Now, say we dump 1 litre of water into this solution. The concentration of our acid will be halved. Plugging this into the equation, we get the concentration of $\ce{H_3O^{+}}$ ions as approx. $0.7 \sqrt{K_a} \ M$. We have two litres of solvent, so we have $1.4 \sqrt{K_a}$ moles of ions, which is considerably higher. So, the numbers all work out, but what's the intuition behind it?