Image of function contains identity elements

Let $U$ be the set of all nonempty subsets of $[0,1]$ that are a union of finitely many closed intervals (where an "interval" that is a single point does not count as an interval). Let $f:Utimes Urightarrow U$ be such that for any $A,Bin U$:

(a) $f(A,B)neq [0,1]$.

(b) $f(A,B)cap Aneqemptyset$ and $f(A,B)cap Bneqemptyset$.

(c) The length (i.e. Lebesgue measure) of $f(X,B)cap A$ is maximized at $X=A$, and the length of $f(A,X)cap B$ is maximized at $X=B$.

Is it true that the image of $f$ must be equal to the set ${Amid f(A,A)=A}$?

Note: The original unsolved question asked whether such $f$ exists. The claim of this question is made in the comments of that question, but the commenter could not recall its proof or whether the proof ever existed.