Suppose that $f,g:[a,b]\rightarrow\mathbb{R}$ are bounded such that $|f(x)-g(x)|\leq\varepsilon$ for each $x\in[a,b]$, for some $\varepsilon>0$.
Show that $|\underline\int_{a}^{b}f-\underline\int_{a}^{b}g|\leq\varepsilon(b-a)$, where \begin{align} \underline\int_{a}^{b}f = \sup_{\mathcal{P}} l(f,\mathcal{P}) \end{align} is the lower (Riemann) integral of $f$ over $[a,b]$.
My attempt based on the hint by Keen-ameteur:
Let $\delta>0$. Pick a partition $\mathcal{P}$ of $[a,b]$ such that \begin{align} \underline\int_{a}^{b}f \leq l(f,\mathcal{P}) + \frac{\delta}{2} \end{align} and \begin{align} \underline\int_{a}^{b}g \leq l(g,\mathcal{P}) + \frac{\delta}{2}. \end{align}
Assume w.l.o.g that $l(f,\mathcal{P}) \geq l(g,\mathcal{P})$. Since $f \leq g + \varepsilon$ we get that \begin{align} |l(f,\mathcal{P}) - l(g,\mathcal{P})| &= l(f,\mathcal{P}) - l(g,\mathcal{P}) \\ &\leq l(g+\varepsilon,\mathcal{P}) - l(g,\mathcal{P}) \\ &=\varepsilon(b-a). \end{align}
The hint I would suggest is using the triangle inequality. i.e., find some sizes $A_\delta,B_\delta$ such that
$$ \Bigg \vert \underline\int_{a}^{b}f-A_\delta\Bigg\vert+ \vert A_\delta-B_\delta\vert + \Bigg\vert B_\delta- \underline\int_{a}^{b}g \Bigg\vert \leq\varepsilon(b-a)+\delta, $$
for all $\delta>0$.
For $A_\delta$ and $B_\delta$, you should have candidates which follow form the definition of $\underline{\int_a^b}$.