drhab has already commented how to informally derive the solution. I will formally prove that this is in fact the correct solution.
To this end, let $Z = \max(X,Y)$. We want to show
$$
E[\min(X,Y)|\sigma(Z)] = \frac{Z}{2} \quad \text{a.s.}
$$
The right hand side is obviously $\sigma(Z)$-measurable. So all we need to show is
$$
E[\min(X,Y)\mathbb{1}_F] = E\left[\frac{Z}{2}\mathbb{1}_F\right]
$$
for all $F \in \sigma(Z)$.
We will show this property for $F = \{Z \leq a \}$ with $a \in \mathbb{R}$. This is enough because these sets form a $\pi$-system (meaning it's $\cap$-stable) generating $\sigma(Z)$ and we can use a dynkin system argument to get the property for all $F \in \sigma(Z)$. Let me know if I should go into more detail here.
Because $X$ and $Y$ are iid $U([0,1])$, $Z$ has the Lebesgue density $f_Z(z) = 2z\mathbb{1}_{[0,1]}(z)$. So we get for the right hand side
$$
E\left[\frac{Z}{2}\mathbb{1}_F\right] = \frac12 \int_{-\infty}^azdP^Z(z) = \int_{-\infty}^a z^2 \mathbb{1}_{[0,1]}(z)dz\\
=
\begin{cases}
0 \quad &a < 0\\
\frac13 a^3 \quad &a \in [0,1]\\
\frac13 \quad &a > 1
\end{cases}.
$$
For the left hand side, we get, using the symmetry of $X$ and $Y$ and $F = \{Z \leq a \} = \{X \leq a \} \cap \{Y \leq a \}$,
$$
E[\min(X,Y)\mathbb{1}_F] = 2E[\min(X,Y)\mathbb{1}_F\mathbb{1}_{\{X \leq Y\}}] = 2E[X \mathbb{1}_{\{X \leq a\}}\mathbb{1}_{\{Y \leq a\}}\mathbb{1}_{\{X \leq Y\}}]\\
= 2\int_0^1\int_0^1 x \mathbb{1}_{\{x \leq a\}} \mathbb{1}_{\{y \leq a\}}\mathbb{1}_{\{x \leq y\}} dy dx\\
= 2 \int_0^1 x \mathbb{1}_{\{x \leq a\}} \int_x^1 \mathbb{1}_{\{y \leq a\}} dy dx.
$$
Now you can distinguish the same $3$ cases as above and notice that you arrive at the same values.
