1. Let the triangle be $\triangle ABC$ and its side-midpoints be $D(0,1),\quad E(1,1),\quad F(1,0).$

2. Recover the vertices from the midpoints.
   For a triangle, $D=\frac{B+C}{2},\qquad E=\frac{C+A}{2},\qquad F=\frac{A+B}{2}.$ Hence, $A=E+F-D,\quad B=F+D-E,\quad C=D+E-F.$

   Now compute: $A=(1,1)+(1,0)-(0,1)=(2,0),$ $B=(1,0)+(0,1)-(1,1)=(0,0),$ $C=(0,1)+(1,1)-(1,0)=(0,2).$

   So the triangle has vertices: $A(2,0),\quad B(0,0),\quad C(0,2).$

3. Find the side lengths.
   Using standard notation: $a=|BC|=2,$ $b=|CA|=\sqrt{(2-0)^2+(0-2)^2}=\sqrt{8}=2\sqrt2,$ $c=|AB|=2.$

4. Use the incentre formula.
   The incentre of $\triangle ABC$ is $\left(\frac{ax_A+bx_B+cx_C}{a+b+c},\; \frac{ay_A+by_B+cy_C}{a+b+c}\right).$

   Therefore its $x$-coordinate is $x_I=\frac{a x_A+b x_B+c x_C}{a+b+c}.$

   Substitute:

   =\frac{4}{4+2\sqrt2}.$$ Simplify: $$x_I=\frac{2}{2+\sqrt2}.$$ Rationalizing, $$x_I=\frac{2(2-\sqrt2)}{(2+\sqrt2)(2-\sqrt2)} =\frac{2(2-\sqrt2)}{4-2} =2-\sqrt2.$$

5. Match with the options.
   $2-\sqrt2$ corresponds to Option B.

Final Answer: $\boxed{2-\sqrt2}$