4. Given a real number $r$, two players, A and B, play the following game.

On the blackboard is written an expression with three absolute values: $S=|\square-\square|+|\square-\square|+|\square-\square|$, where the 6 blanks “$\square$” are yet to be filled with numbers. In each turn, A selects a real number from the interval $[0,1]$ (A can choose the same number in different turns), and B fills this number into one of the blanks. After six turns, all 6 blanks are filled, and the value of $S$ is determined. If $S \geqslant r$, then A wins; otherwise, B wins. Find all real numbers $r$ such that A has a winning strategy.
First, discuss the case of 4 blanks: $T=|\square-\square|+|\square-\square|$.
Player A has a strategy to ensure $T \geqslant \frac{3}{2}$: Player A first chooses 0 (choosing 1 is also possible), and Player B's first move is meaningless, $T=|0-\square|+|\square-\square|$. Player A then chooses 1. If Player B fills it in the same absolute value as 0, Player A can sequentially choose 0, 1, making $T=2$; if Player B fills it in the other absolute value, $T=|0-\square|+|1-\square|$, Player A then chooses $\frac{1}{2}$, which will result in $\frac{1}{2}$ in one of the absolute values. The last number can be chosen by Player A to make the other absolute value 1, thus $T=\frac{3}{2}$.
Player B has a strategy to ensure $T \leqslant \frac{3}{2}$: If the first two numbers chosen by Player A differ by no more than $\frac{1}{2}$, Player B can fill them in the same absolute value, making one absolute value no more than $\frac{1}{2}$, and the other no more than 1, thus $T \leqslant \frac{3}{2}$; if the first two numbers chosen by Player A differ by more than $\frac{1}{2}$, let them be $a, b$, and $b-a>\frac{1}{2}$, then Player B fills them in different absolute values, set $T=|a-\square|+|b-\square|$. It is easy to see that $a \in\left[0, \frac{1}{2}\right], b \in\left[\frac{1}{2}, 1\right]$, then when Player A chooses the third number $c \in\left[0, \frac{1}{2}\right]$, it can ensure $|a-c| \leqslant \frac{1}{2}$, and when $c \in\left[\frac{1}{2}, 1\right]$, it can ensure $|b-c| \leqslant \frac{1}{2}$, and the other absolute value is no more than 1, thus $T \leqslant \frac{3}{2}$.
Returning to the original problem. Player A has a strategy to ensure $S \geqslant \frac{15}{8}$: Player A sequentially chooses 0, 1, if Player B fills them in the same absolute value, by the discussion of $T$, Player A can make $S \geqslant 1+\frac{3}{2}=\frac{5}{2}>\frac{15}{8}$;
Assume Player B fills 0, 1 in different absolute values, Player A then chooses $\frac{3}{8}$, if Player B fills $\frac{3}{8}$ in the same absolute value as 0 or 1, by the discussion of $T$, Player A can make $S \geqslant \frac{3}{8}+\frac{3}{2}=\frac{15}{8}$;
The following discusses Player B filling $\frac{3}{8}$ in the third absolute value, i.e., $S=|0-\square|+|1-\square|+\left|\frac{3}{8}-\square\right|$. Player A chooses $\frac{6}{8}$, if Player B places it in the first absolute value, then Player A chooses 0, 0, getting $S \geqslant \frac{6}{8}+1+\frac{3}{8}=\frac{17}{8}>\frac{15}{8}$; if Player B places it in the second absolute value, then Player A chooses 1, 1, getting $S \geqslant 1+\frac{2}{8}+\frac{5}{8}=\frac{15}{8}$;
If Player B places it in the third absolute value, by the discussion of $T$, Player A can make $S \geqslant \frac{3}{2}+\frac{3}{8}=\frac{15}{8}$. Player B has a strategy to ensure $S \leqslant \frac{15}{8}$: If the difference between the first two numbers chosen by Player A is no more than $\frac{3}{8}$, or the third number chosen by Player A differs from one of the first two numbers by no more than $\frac{3}{8}$, then Player B can fill the two numbers with a difference of no more than $\frac{3}{8}$ in the same absolute value within the first three rounds, by the discussion of $T$, Player B can make $S \leqslant \frac{3}{8}+\frac{3}{2}=\frac{15}{8}$; if the first three numbers chosen by Player A have differences greater than $\frac{3}{8}$, then Player B fills these three numbers in different absolute values, assume $S=|a-\square|+|b-\square|+|c-\square|$, where $b-a>\frac{3}{8}, c-b>\frac{3}{8}$, by symmetry, assume $b \leqslant \frac{1}{2}$. Let the fourth number chosen by Player A be $d$.
Case 1: $d \in\left[c-\frac{2}{8}, 1\right]$. Player B places $d$ and $c$ in the same absolute value, since $c>\frac{6}{8}$, then $|c-d| \leqslant \frac{2}{8}$, and the sum of the first two absolute values is no more than $1-a+1-b=2-(a+b) \leqslant 2-\frac{3}{8}=\frac{13}{8}$, thus $S \leqslant \frac{13}{8}+\frac{2}{8}=\frac{15}{8}$;
Case 2: $d \in\left[b-\frac{3}{8}, b+\frac{3}{8}\right]$. Player B places $d$ and $b$ in the same absolute value, then $|b-d| \leqslant \frac{3}{8}$. Since $a \in\left[0, \frac{1}{2}\right], c \in\left[\frac{1}{2}, 1\right]$, by the discussion of $T$, Player B can make $S \leqslant \frac{3}{2}+\frac{3}{8}=\frac{15}{8}$;
Case 3: $d \in\left[0, a+\frac{3}{8}\right]$. Player B places $d$ and $a$ in the same absolute value, since $a<\frac{2}{8}$, then $|a-d| \leqslant \frac{3}{8}$. Since $b \in\left[0, \frac{1}{2}\right], c \in\left[\frac{1}{2}, 1\right]$, by the discussion of $T$, Player B can make $S \leqslant \frac{3}{8}+\frac{3}{2}=\frac{15}{8}$. Notice that $\left[0, a+\frac{3}{8}\right] \cup\left[b-\frac{3}{8}, b+\frac{3}{8}\right] \cup\left[c-\frac{2}{8}, 1\right]=[0,1]$, and $\frac{3}{8}<b \leqslant \frac{1}{2}$, thus the above three cases cover all possibilities for $d$.
In summary, the $r$ for which Player A has a winning strategy is all real numbers no greater than $\frac{15}{8}$.