The answer is no. Suppose that it is possible. We assert that the previous triangle $A_{n-1}B_{n-1}C_{n-1}$ has rational coordinates. In fact, the points $A_{n-1}$, $B_{n-1}$, $C_{n-1}$ are the feet of the altitudes of the triangle $A_nB_nC_n$. Therefore it is enough to show that, if a line segment has its ends with rational coordinates, then the foot of the perpendicular line passing through a point of the plane with rational coordinates has also rational coordinates. This really happens because the coordinates $(x, y)$ of the foot of the perpendicular are the solutions of the system $y = ax + b$, $y = -\frac{1}{a}x + c$ with $a, b, c$ rational. Therefore, every time in the previous step the coordinates must be rational and so, we arrive to the conclusion that the coordinates of the triangle $A_0B_0C_0$ must be rational. Then from the area formula using coordinates of the vertices we find that the area of the triangle is a rational number. This contradicts the supposition that the area of the triangle is equal to $\sqrt{2}$.