Solution:

For every real $x$, $$f(1-x)=|\lfloor 1-x\rfloor|-|\lfloor x\rfloor|=-f(x).$$

Equivalently, replacing $x$ by $\frac{1}{2}+t$ gives $f\!\left(\frac{1}{2}-t\right)=-f\!\left(\frac{1}{2}+t\right)$. This is point symmetry about $\left(\frac{1}{2},0\right)$.

Thus, D is the correct answer.