How do I show that $\int_{-\infty}^\infty \frac{ \sin x \sin nx}{x^2} \ dx = \pi$?

It depends how rigorous you want to be. For $n=1$ this is a classic integral, that I'll assume you have seen before/can easily find. For $n>1$ we have the following generalization if we let $a>b\geqslant 0$

$$\begin{aligned}2\int_0^{\infty}\frac{\sin(ax)\sin(bx)}{x^2}\;dx &=\int_0^{\infty}\frac{\cos((a-b)x)-\cos((a+b)x)}{x^2}\;dx \\ &= \int_0^{\infty}\int_{a-b}^{a+b}\frac{\sin(xy)}{x}\;dy \;dx\\ &=\int_{a-b}^{a+b}\int_0^{\infty}\frac{\sin(xy)}{x}\;dx \;dy \\ &=\int_{a-b}^{a+b}\frac{\pi}{2}\;dy \\ &=\pi b\end{aligned}$$

Another way to see why it should be so is to go to the frequency domain. Let $f_a(x)=\frac{\sin ax}x$ and $a\ge b>0\,$. The Fourier transform of $f_a(x)$ is a step: $$ F[f_a](\xi)=\sqrt{\frac\pi2}\theta(a-|x|), $$ there $\theta$ is the Heaviside step function. By the properties of Fourier transform we have $$ \int_{-\infty}^{\infty}f_a(x)f_b(x)\,dx=F[f_af_b](0)= F[f_a]*F[f_b](0)= $$ $$\int_{-\infty}^{\infty}F[f_a](\xi)F[f_b](-\xi)\,d\xi= \frac\pi2 \int_{-b}^b d\xi=\pi b. $$

Another approach is integration by parts:

$$ \begin{align} \int_{-\infty}^\infty \frac{ \sin x \sin nx}{x^2} \mathrm dx &= \int_{-\infty}^\infty\frac{\cos x\sin nx+n\sin x\cos nx}x\mathrm dx \\ &= \frac12\int_{-\infty}^\infty\frac{\sin(n+1)x+\sin(n-1)x+n(\sin(n+1)x-\sin(n-1)x)}x\mathrm dx \\ &= \frac12(1+1+n-n)\int_{-\infty}^\infty\frac{\sin x}x\mathrm dx \\ &= \pi\;. \end{align} $$

For $n=1$, the terms with $\sin(n-1)x$ don't occur, but the result is the same.

Let $n\ge 1$. You can integrate $$f(z) = \frac{\sin(z)e^{inz}}{z^2}$$ around a big half-disc $U_R$ in the upper half-plane. The integral over the circle-part will go to $0$ for $R\to\infty$ (that's where $n\ge 1$ is needed). Therefore

$$\int_{-\infty}^\infty \frac{\sin(x)\sin(nx)}{x^2}\, dx = \lim_{R\to \infty} \mathrm{Im}\left[\oint_{\partial U_R} f(z) \, dz\right] = \mathrm{Im}\left[\pi i \;\mathrm{Res}_{z=0}(f(z))\right] = \pi$$