= Solution
Work on $([0,1],\mathcal B,dt)$ as a <probability space>, with <Lebesgue measure> of total mass one. Let $\mathcal F_n$ be the <filtration> generated by the level-$n$ half-open dyadic cells together with the separate null cell $\{1\}$. Define the <dyadic slope martingale> by
$$
G_n(t)=2^n\bigl[f((k+1)2^{-n})-f(k2^{-n})\bigr]
\quad\text{on }[k2^{-n},(k+1)2^{-n}).
$$
The endpoint $t=1$ may be assigned any value, since it is a null set for <Lebesgue measure>. The two child slopes average to their parent slope, by telescoping the two increments. Therefore $\mathbb E[G_{n+1}\mid\mathcal F_n]=G_n$ <almost everywhere>, so this is a <martingale>. The <Lipschitz condition> gives $|G_n|\leq K$ everywhere except possibly at the freely chosen endpoint, where we take zero.
Apply the <Lp martingale convergence theorem> with $p=2$. Its limit $g$ has $|g|\leq K$ <almost everywhere> and $G_n\to g$ in <L1 norm> as well, by <Cauchy-Schwarz inequality>. Choose a measurable representative of $g$ and set it to zero on any exceptional null set; it is then a bounded <measurable function> on the entire interval.
Set $f_n(x)=f(0)+\int_0^xG_n(t)dt$. Telescoping at the grid points shows that $f_n$ is the <linear interpolation> of $f$ on the dyadic grid. The <Lipschitz condition> gives $\|f_n-f\|_\infty\leq K2^{-n}\to0$. Also
$$
\sup_{0\leq x\leq1}\left|\int_0^x(G_n-g)(t)dt\right|\leq\|G_n-g\|_1\to0.
$$
The two uniform limits coincide, giving the <absolutely continuous function> representation
$$
\boxed{f(x)=f(0)+\int_0^xg(t)dt\quad(0\leq x\leq1),\qquad |g(t)|\leq K.}
$$
The chosen bound holds for every $t$ after the null-set modification; the <integral> identity holds for every $x$ simultaneously.
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