You can succeed in proving $\frac{\sin x}{x}$ is not Lebesgue integrable over $[0,\infty[$ also by using some street-fighting mathematics.
Actually, you only have to show that: $\int_\pi^\infty \frac{|\sin x|}{x}\ \text{d} x =\infty\; ,$ for the integral $\int_0^\pi \frac{|\sin x|}{x}\ \text{d}x$ is finite (due to $\lim_{x\to 0^+} \frac{\sin x}{x} =1$ and continuity of $\frac{\sin x}{x}$ in $]0,\pi]$).
Let $f(x):=\frac{|\sin x|}{x}$ for sake of simplicity. Then $f$ is nonnegative and concave in each interval of the type $[k\pi, (k+1)\pi]$ and it attains its global minimum (i.e. $0$) in $x_k:=k\pi$, with $k\in \mathbb{N}$; moreover, $f$ attains local maximum in $\xi_k \in ]k\pi,(k+1)\pi[$, where $\xi_k$ is the unique solution of: $\sin x=x\ \cos x$ in $[k\pi,(k+1)\pi]$.
The triangle $\mathfrak{T}_k$ having vertices in $A_k:=(k\pi ,0)$, $B_k:=((k+1)\pi ,0)$ and $C_k:=(\xi_k,f(\xi_k))$ lies in the trapezoid $\mathfrak{R}_k:=\{(x,y)\in \mathbb{R}^2|\ k\pi\leq x\leq (k+1)\pi,\ 0\leq y\leq f(x)\}$ by concavity, hence for each index $k$: $\int_{k\pi}^{(k+1)\pi} f(x)\ \text{d} x =\operatorname{Area}(\mathfrak{R}_k) \geq \operatorname{Area}(\mathfrak{T}_k)=\frac{\pi}{2}\ f(\xi_k)$ and: $\tag{1} \int_{\pi}^{(k+1)\pi} f(x)\ \text{d} x\geq \sum_{n=1}^k \frac{\pi}{2}\ f(\xi_n)\; .$
Now, you win if you prove that the RHside of (1) is the $k$-th partial sum of a positively divergent series.
You can prove that: $\xi_k = \frac{\pi}{2} +k\pi -\varepsilon_k = \frac{\pi}{2} (2k+1)-\varepsilon_k$ with $0<\varepsilon_k<\pi/2$ and $\varepsilon \to 0$ as $k\to \infty$ (cfr. Mahajan, Street-fighting Mathematics, 6.4), thus you get: $\begin{split} f(\xi_k) & = \frac{|\sin \xi_k|}{\xi_k} \\ &= \frac{|\sin (\pi/2 +k\pi -\varepsilon_k)|}{\frac{\pi}{2} (2k+1)-\varepsilon_k} \\ &= \frac{\sin (\pi/2 -\varepsilon_k)}{\frac{\pi}{2} (2k+1)-\varepsilon_k} &\qquad \text{(} \sin t \text{ is periodic)} \\ &\geq \frac{2}{\pi}\ \frac{\sin (\pi/2 -\varepsilon_k)}{2k+1} &\qquad \text{(denominator increased + algebra)} \\ & = \frac{2}{\pi}\ \frac{\cos \varepsilon_k}{2k+1} &\qquad \text{(trigonometric trick)} \\ & \geq \frac{2}{\pi}\ \frac{1- \frac{1}{2} \varepsilon_k^2}{2k+1}\; , \end{split}$ and the latter inequality holds because of the elementary inequality $\cos t \geq 1-\frac{1}{2}\ t^2$. Therefore you find: $\sum_{n=1}^k \frac{\pi}{2}\ f(\xi_n) \geq \sum_{n=1}^k \frac{1- \frac{1}{2} \varepsilon_n^2}{2n+1}$ and the RHside diverges in the positive sense when $k$ goes to $\infty$ (for the summand $\frac{1- \frac{1}{2} \varepsilon_n^2}{2n+1}$ is asymptotically equivalent to that of a harmonic series). Finally you can pass to the limit in (1) to get $\int_\pi^\infty f(x)\ \text{d} x =\infty $ as you claimed.