Answer
See explanations.
Work Step by Step
Step 1. We assume $\vec u=\langle u_x, u_y \rangle$ and $\vec v=\langle v_x, v_y \rangle$
Step 2. Use the projection formula, $proj_v\vec u=(\frac{\vec u\cdot\vec v}{|\vec v|^2})\vec v=(\frac{u_xv_x+u_yv_y}{v_x^2+v_y^2}
)\langle v_x, v_y \rangle$
Step 3. Calculate vector difference $\vec u-proj_v\vec u=\langle u_x-(\frac{u_xv_x+u_yv_y}{v_x^2+v_y^2})v_x, u_y-(\frac{u_xv_x+u_yv_y}{v_x^2+v_y^2})v_y \rangle$
Step 4. Calculate the dot product of the above two vectors: $(proj_v\vec u)\cdot(\vec u-proj_v\vec u)
=(\frac{u_xv_x+u_yv_y}{v_x^2+v_y^2})(u_xv_x-(\frac{u_xv_x+u_yv_y}{v_x^2+v_y^2})v_x^2+u_yv_y-(\frac{u_xv_x+u_yv_y}{v_x^2+v_y^2})v_y^2)=0$ (note that the denominator $v_x^2+v_y^2$ will be cancelled after combining the second and fourth terms)
Step 5. We proved that the two vectors are orthogonal as their dot product is zero.
