For two independent $U(0,1)$ random variables, the joint density of the maximum and minimum is uniformly distributed on the triangular region with vertices at $(0,0)$, $(1,0)$ and $(1,1)$. So, yes, given that the maximum has value $k \in (0,1)$, the minimum is uniformly distributed on $(0,k]$. Similarly, given that the minimum has value $\ell \in (0,1)$, the maximum is uniformly distributed on $[\ell, 1)$.
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Let $W = \max\{X,Y\}$ and $Z = \min\{X,Y\}$. Then, for $a \geq b$, $\begin{align} F_{W,Z}(a,b) &= P\{W \leq a, Z \leq b\}\\ &= P\left(\{W \leq a\}\cap \{Z \leq b\}\right)\\ &= P\left(\{X \leq a, Y \leq b\} \cup \{X \leq b, Y \leq a\}\right)\\ &= P\{X \leq a, Y \leq b\} + P\{X \leq b, Y \leq a\} - P\{X \leq b, Y \leq b\} )\\ &= F_{X,Y}(a,b) + F_{X,Y}(b,a) - F_{X,Y}(b,b). \end{align}$ If one of the steps puzzles you, think about $P(A\cup B) = P(A) + P(B) - P(A\cap B)$). On the other hand, if $a < b$, then $F_{W,Z}(a,b) = P\{X \leq a, Y \leq a\} = F_{X,Y}(a,a).$ For jointly continuous random variables $X$ and $Y$, this gives (upon taking partial derivatives with respect to $a$ and $b$ that $f_{W,Z}(a,b) = \begin{cases} f_{X,Y}(a,b) + f_{X,Y}(b,a), &\text{if}~ a \geq b,\\ 0, &\text{if}~ a < b. \end{cases}$ If you think of the joint density $f_{X,Y}$ as a solid sitting on the plane, just fold it over the diagonal line $a=b$ to get the joint density $f_{\max,\min}$ of the maximum and the minimum. For your instance, the unit cube becomes a right triangular prism of height $2$ with vertices at $(0,0)$, $(1,0)$ and $(1,1)$. Thus, the conditional density of one of the two random variables given the value of the other, being just a scaled version of the cross-section of the joint density, is a uniform density.