To answer this, imagine a simple case of induction.
Note that this is going to be a non-rigorous, non mathematical answer; but I believe it should serve the purpose.
Imagine a metal sphere placed a little away from a positive point charge (say $q$), which happens to be fixed. Now, assuming the electron-sea model for the metal sphere (only because it makes things a little easier to visualise), the charge $q$ pulls the electrons towards its nearer side, pushing away the remaining positive metal 'kernels' on the farther side. These are your 'induced' charges. Call the distances from the negative and positive charge densities $d-$ and $d+$; $(d+)>(d-)$.
Now, the induced charges $q_{in}$ interact wit our fixed charge $q$, and we have attractive force between $q$ and $-q_{in}$, say $F_{att}$; and a repulsive force $F_{rep}$ between $q$ and $+q_{in}$. Now, according to coulomb's law; both these forces vary inversely with the square of the distance; in other words, the closer the charges, the stronger the force. Since $(d{-})<(d+)$,i.e. the negative charge density happens to be closer, and hence $F_{att}>F_{rep}$.
So the net force on your (still) neutral sphere( this is important-the sphere is still neutral, as it should be-conservation of charge) is attractive, and hence a positive charge attracts a neutral body.
On the other hand, if the force law was distance independent, then the size of $d+$ and $d-$ would be irrelevant; both would experience equal forces(note that the magnitude of induced charges is same); or $F_{att}=F_{rep}$ in this case. Hence, your sphere would be in equilibrium.
NOTE-1) The sphere remains neutral throughout because the induced charges $-q_{in}$ and $+q_{in}$ exactly cancel out. This is in accordance of conservation of charge, or more fundamentally, the neutrality of atoms.