Intuition of joint density of min(X,Y) and max(X,Y)

The problem with the joint distribution of $U$ and $Y$ is that you do not know which of $X$ and $Y$ is $U$ and which is $V$. Similarly the difficulty with $P(U \le u, V \le v) $, i.e. at least one of $X$ and $Y$ is below $u$ and both below $v$, is handling the question of which one is below $u$ (taking into account that they both may be).

The trick in the book solution is spotting that $$P(U \le u, V \le v) = P(V\le v)-P(u \lt U \le v, V\le v)$$ gives you two terms which each involve $X$ and $Y$ being in the same interval.

It is in fact easy to go the naive route: with $0 \le u \le v$, you could say $$P(U \le u, V \le v) = P(X \le u, u \lt Y \le v)+ P( u \lt X \le v, Y \le u) + P(X \le u, Y \le u) $$ $$= P(X \le u)P( u \lt Y \le v)+ P( u \lt X \le v)P(Y \le u) + P(X \le u)P( Y \le u)$$ by independence and derive the answer that way. You will get the same answer since this is obviously equal to $$P(X \le v)P(Y \le v) - P(u \lt X \le v)P( u \lt Y \le v).$$

Rewriting it like this guarantees that both random variables $X$ and $Y$ are wedged between two numbers, and the probability that this happens can be calculated from their distribution.

In particular, $\max (X,Y)\le v$ guarantees that both $X$ and $Y$ are not larger than $v$, while $u<\min(X,Y)\le\max(X,Y)\le v$ guarantees that $X$ and $Y$ lie both between $u$ and $v$. But the probability that this is the case factorizes because $X$ and $Y$ are independent by assumption.

Therefore, splitting the probabilities like that allows to employ the independence of $X$ and $Y$, while $U$ and $V$ are not independent.