1. Assume the contrary: Suppose the union of the rectangles $R_1, R_2, R_3$ does not cover all points inside the triangle $ABC$. This means there exists at least one point inside $ABC$ that is not covered by any of the rectangles.

2. Covering the perimeter: Since the union of the rectangles covers the perimeter of $ABC$, every point on the sides $AB, BC,$ and $CA$ is contained in at least one of the rectangles.

3. Finite region bounded by rectangles: Consider the region inside $ABC$ that is not covered by the rectangles. This region must be bounded by the sides of the rectangles and the sides of the triangle $ABC$. Since the rectangles have sides parallel to two fixed perpendicular directions, the uncovered region must be a polygon with sides parallel to these directions.

4. Shape of the uncovered region: The uncovered region cannot be a triangle because a triangle cannot have all its sides parallel to two fixed perpendicular directions. Instead, the uncovered region must be a quadrilateral or a more complex polygon.

5. Contradiction with infinite region: If the uncovered region is a quadrilateral, then two of its sides must be parallel to one direction and the other two sides must be parallel to the perpendicular direction. This configuration would imply that the uncovered region extends infinitely in at least one direction, which contradicts the fact that the region is bounded within the finite area of triangle $ABC$.

6. Conclusion: Since the assumption that there exists an uncovered point inside $ABC$ leads to a contradiction, we conclude that the union of the rectangles $R_1, R_2, R_3$ must cover all points inside the triangle $ABC$.

$\blacksquare$