Show that $\int^{\infty}_{0}\left(\frac{\sin(x)}{x}\right)^2 < 2$

Well, this likely isn't what you had in mind, but you could just evaluate the integral. In this case, Parseval-Plancherel's theorem works:

$$\int_{-\infty}^{\infty} dx\, |f(x)|^2 = \frac{1}{2 \pi}\int_{-\infty}^{\infty} dk\, |\hat{f}(k)|^2$$

where $\hat{f}$ is the Fourier transform of $f$. For $f(x)=\sin{x}/x$, we have

$$\int_{-\infty}^{\infty} dx\, \left ( \frac{\sin{x}}{x}\right)^2 = \frac{1}{2 \pi} \int_{-1}^1 dk \, \pi^2 = \pi$$

so that

$$\int_0^{\infty} dx\, \left ( \frac{\sin{x}}{x}\right)^2 = \frac{\pi}{2} < 2$$

Hint: $$\lim_{x\to0}\frac{\sin x}{x}=1.$$

By the Laplace transform (since $\mathcal{L}(\sin^2 x)=\frac{2}{s(4+s^2)}$ and $\mathcal{L}^{-1}\left(\frac{1}{x^2}\right)=s$) $$ I=\int_{0}^{+\infty}\frac{\sin^2 x}{x^2}\,dx = \int_{0}^{+\infty}\frac{2\,ds}{4+s^2}\stackrel{s\mapsto 2t}{=} \int_{0}^{+\infty}\frac{dt}{1+t^2}=\color{red}{\frac{\pi}{2}}$$ and with a more elementary approach, $\left|\sin(x)\right|\leq\min(|x|,1)$ implies: $$ I \leq \int_{0}^{1}\frac{x^2}{x^2}\,dx + \int_{1}^{+\infty}\frac{1}{x^2}\,dx = 2.$$