After proving the upper bound, give a piecewise-linear function that attains it

Use the contraction condition twice on each side of $f(300)=f(900)$

Applying the contraction inequality twice gives $$\begin{aligned} &|f(f(400))-f(f(300))|\\ &\quad\le\frac12|f(400)-f(300)|\\ &\quad\le25, \end{aligned}$$ and similarly $$|f(f(800))-f(f(900))|\le25.$$

Set $M=f(f(300))=f(f(900)).$ The triangle inequality now yields $$\begin{aligned} &|f(f(800))-f(f(400))|\\ &\quad\le|f(f(800))-M|\\ &\qquad+|M-f(f(400))|\\ &\quad\le50. \end{aligned}$$

To attain the bound, define $f$ by linear interpolation through the points $$\begin{gathered}(300,600),(400,550),(550,575),\\ (650,625),(800,650),(900,600).\end{gathered}$$ and set $f(x)=600$ for $x\leq300$ or $x\geq900.$ Every segment has slope with absolute value at most $\dfrac12,$ so the contraction condition holds. In particular, $f(300)=f(900)=600,$ $f(400)=550,$ and $f(800)=650.$ Hence $f(f(400))=f(550)=575,$ while $f(f(800))=f(650)=625,$ giving the difference $50.$

Thus, the answer is B .