How to prove absolute summability of sinc function?

It doesn't. Using the convexity of $1/x$,

$$\int_0^\infty \left\vert\frac{\sin x}{x}\right\vert \,\mathrm{d}x=\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\left\vert\frac{\sin x}{x}\right\vert \,\mathrm{d}x>\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\frac{\left\vert\sin x\right\vert}{(k+1/2)\pi} \,\mathrm{d}x=\frac{2}{\pi}\sum_{k=0}^\infty\frac{1}{k+1/2}\;,$$

which diverges since the harmonic series diverges.

It doesn't. This function is a sequence of bumps of decreasing size. The $n^{\text{th}}$ "bump" bounds area on the order of $\frac{1}{n}$ (there are a number of ways to show this), but $\sum_{n=1}^\infty \frac{1}{n} = \infty$.

$\int_0^\infty \frac{\sin x}{x} dx$ is in fact my most favorite example of a function which is Riemann integrable but not Lesbegue integrable. (just to tease people which think that Lesbegue is so much better than Riemann; in fact this integral might appear more frequent in applications than any integral which is Lesbegue but not Riemann integrable)