If $1\leq a<b$, then $$ \left|\int_{a}^{b}\frac{\sin(x)}{x}dx\right|\leq 3.$$
Proceeding by integration by parts; let $u(x)=\sin(x)$ and $dv(x)=1/x$, then $u'=\cos(x)$ & $v(x)=\log(x)$. We get, $$\bigg|\sin(x)\log(x)\bigg|_a^b-\int_a^b \cos(x)\log(x)dx\bigg|$$ Now, i think it might be convenient to use the fact that $sine$ and $cosine$ are bounded above by 1 and $\log(x)\leq x-1$. Thanks
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The stationary points of the sine integral function $\text{Si}(x)$ occur at $x\in\pi\mathbb{Z}\setminus\{0\}$ by the fundamental theorem of calculus, hence the function $\text{Si}(x)$ is bounded between $\text{Si}(1)$ and $\text{Si}(\pi)$ on the interval $(1,+\infty)$.
It follows that we just have to prove: $$ \int_{1}^{\pi}\frac{\sin x}{x}\,dx \leq 3 $$ or: $$ \int_{0}^{\pi-1}\frac{\sin x}{\pi-x}\,dx \leq 3.$$ However, that is trivial, since the LHS is bounded by: $$ \int_{0}^{\pi-1}\frac{4x}{\pi^2}\,dx = \frac{2(\pi-1)^2}{\pi^2} = \color{red}{0.9294\ldots} $$
A weaker but easier inequality follows from $ \int_{1}^{\pi}\frac{\sin x}{x}\,dx \leq \int_{1}^{\pi} 1\,dx = \pi-1 < 3.$
Your approach also works. By integration by parts: $$ \text{Si}(b)-\text{Si}(a) = \left.\frac{1-\cos x}{x}\right|_{a}^{b}+\int_{a}^{b}\frac{1-\cos x}{x^2}\,dx $$ but since $0\leq (1-\cos x)\leq 2$, $$ \left|\text{Si}(b)-\text{Si}(a)\right|\leq \max_{x\geq 1}\frac{1-\cos x}{x}+\int_{1}^{+\infty}\frac{2}{x^2}\,dx < 3. $$
Actually, you can improve the result. Noting $$ \int\frac{\sin x}{x}dx=-\int\frac{1}{x}d\cos x=-\frac{\cos x}{x}+\int\frac{\cos x}{x^2}dx $$ one has \begin{eqnarray} \bigg|\int_a^b\frac{\sin x}{x}dx\bigg|&=&\bigg|-\frac{\cos x}{x}|_a^b+\int_a^b\frac{\cos x}{x^2}dx\bigg|\\ &\le&\bigg|-\frac{\cos b}{b}+\frac{\cos a}{a}\bigg|+\int_a^b\frac{1}{x^2}dx\\ &\le&\frac{1}{a}+\frac{1}{b}-\frac{1}{x}\bigg|_a^b\\ &\le&\frac{1}{a}+\frac{1}{b}+\frac{1}{a}-\frac{1}{b}\\ &=&\frac{2}{a}\\ &\le&2. \end{eqnarray}