Proof of $\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$

Well, it's not hard to reduce this integral to $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx$: Just integrate by parts in $\displaystyle \int_0^{\infty} {\sin^2(x) \over x^2}\,dx$, integrating $\displaystyle {1 \over x^2}$ and differentiating $\displaystyle \sin^2(x)$. You're left with $\displaystyle \int_0^{\infty} {\sin(2x) \over x}\,dx$ which reduces to the $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx$ integral after changing variables from $\displaystyle x$ to $\displaystyle 2x$.

So any elementary proof that $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx = {\pi \over 2}$ is effectively also an elementary proof that $\displaystyle \int_0^{\infty} {\sin^2(x) \over x^2}\,dx$ is also $\displaystyle {\pi \over 2}$.

Let $f(x)=\max\{0,1-|x|\}$. It is easy to calculate the Fourier transform $$\hat{f}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-ix\xi}dx=\left(\frac{\sin(\xi/2)}{\xi/2}\right)^2.$$ Taking the inverse Fourier transform, we get $$\int_{-\infty}^{\infty}\left(\frac{\sin(\xi/2)}{\xi/2}\right)^2e^{ix\xi}d\xi=2\pi f(x),$$ and the result follows.

The second integral can be computed in a similar way. Just take $f(x)=\chi_{[-1,1]}(x)$ (the indicator function of the interval $[-1,1]$).

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Edit. It might be interesting to note that there are analogous formulas for the sinc sums $$\sum_{n=1}^{\infty}\frac{\sin n}{n}=\sum_{n=1}^{\infty}\left(\frac{\sin n}{n}\right)^2= \frac{\pi}{2}-\frac{1}{2}.$$

I learned about this from the note "Surprising Sinc Sums and Integrals" by Baillie, Borwein, and Borwein (can be found through a quick web search).

More generally, there is a result due to Wolstenholme (I can't find a link) that says $$ \int_0^\infty \left( \frac {\sin x}{x} \right)^n dx = \frac{1}{(n-1)!} \frac{\pi}{2^n} \left\lbrace n^{n-1} - { n \choose 1 } (n-2)^{n-1} + { n \choose 2 } (n-4)^{n-1} - \cdots \right\rbrace .$$