1. We are given the homogeneous system $(A^2B^2 - B^2A^2)X = O$ where $A,B$ are $3\times 3$ real matrices, $A$ is symmetric and $B$ is skew-symmetric.

2. First, note the transpose properties:

   • Since $A$ is symmetric, $A^T=A$.
   • Since $B$ is skew-symmetric, $B^T=-B$.

3. Now examine $A^2$ and $B^2$.

   For $A^2$: $(A^2)^T = (AA)^T = A^TA^T = AA = A^2$ so $A^2$ is symmetric.

   For $B^2$: $(B^2)^T = (BB)^T = B^TB^T = (-B)(-B)=B^2$ so $B^2$ is also symmetric.

4. Let $M = A^2B^2 - B^2A^2.$ We find the transpose of $M$: $M^T = (A^2B^2 - B^2A^2)^T$ $= (B^2)^T(A^2)^T - (A^2)^T(B^2)^T$ $= B^2A^2 - A^2B^2$ $= -(A^2B^2 - B^2A^2)$ $= -M.$ Hence $M$ is skew-symmetric.

5. A standard fact: every real skew-symmetric matrix of odd order has determinant zero.

   Since $M$ is a $3\times 3$ skew-symmetric matrix, $\det(M)=0.$

   Proof briefly: $\det(M)=\det(M^T)=\det(-M)=(-1)^3\det(M)=-\det(M),$ so $\det(M)= -\det(M) \implies \det(M)=0.$

6. Therefore the coefficient matrix $M$ is singular, so the homogeneous system $MX=O$ has non-trivial solutions.

   Any homogeneous system always has the trivial solution $X=O$, and since $M$ is singular, it has at least one free variable. Hence it has infinitely many solutions.

7. Now check the options:

   • A: no solution — impossible for a homogeneous system.
   • B: exactly two solutions — impossible over real numbers for a linear homogeneous system.
   • C: infinitely many solutions — correct.
   • D: a unique solution — would require $\det(M)\neq 0$, false.

Therefore, the correct option is $\boxed{\text{C}}.$