show that $\intop_{0}^{1}\frac{\sin^{2}\left(x\right)}{x^{2}}$ diverges

It should not diverge. $\frac{\sin^2x}{x^2}$ is bounded over $[0,1]$, which means that its integral over the same range is bounded.

When you split the integral into $\int\frac1{x^2}-\int\frac{\cos^2x}{x^2}$, both of those integrals do diverge, so you created an $\infty-\infty$ scenario, which is nonsense.