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\left| x - \left| x-1 \right| \right| = \lfloor x \rfloor

\(\left| x - \left| x-1 \right| \right| = \lfloor x \rfloor \)

firstly

\(\left| x - \left| x-1 \right| \right| \ge 0\qquad \text{because it is an absolute value}\\ so \;\;\;\lfloor x \rfloor \ge0\\ \therefore\quad x\ge0 \)

so

\(\left| x - ( x-1 ) \right| = \lfloor x \rfloor\\ \left| x - x+1 \right| = \lfloor x \rfloor\\ 1 = \lfloor x \rfloor\\ 1\le x<2\)