Solution:

I claim the answer is $\frac{4}{9}$. Let $s = x + y$, $p = x y$, so $x$ and $y$ are $\frac{s \pm \sqrt{s^{2} - 4p}}{2}$. Since $x$ and $y$ are real, $s^{2} - 4p \geq 0$.

If one of the three quantities is less than or equal to $\frac{1}{9}$, then at least one of the others is at least $\frac{4}{9}$ by the pigeonhole principle since they add up to $1$.

Assume that $s - 2p < \frac{4}{9}$, then $s^{2} - 4p < (\frac{4}{9} + 2p)^{2} - 4p$, and since the left side is non-negative we get
$$0 \leq p^{2} - \frac{5}{9} p + \frac{4}{81} = \left(p - \frac{1}{9}\right)\left(p - \frac{4}{9}\right).$$
This implies that either $p \leq \frac{1}{9}$ or $p \geq \frac{4}{9}$, and either way we're done.

This minimum is achieved if $x$ and $y$ are both $\frac{1}{3}$, so the answer is $\frac{4}{9}$, as claimed.