Define the premeasure $\mu_f$ on the algebra of half-open intervals $(a,b]$ by $\mu_f((a,b])=f(b)-f(a).$
Now, $\mu_f$ extends $uniquely$ to a measure $\mu$ on $\mathscr B(\mathbb R).$ Next, define, for each $A\in \mathscr B(\mathbb R)$, $\nu (A)=\int_A f'd\lambda. \ $ Since $\nu$ also extends $\mu_f,\ $ we have by uniqueness that $\nu=\mu$ and absolute continuity follows immediately, and clearly $\frac{d\mu}{d\lambda}=f'$.
edit: I think I have a proof from scratch:
The Lebesgue-Stieljes measure $\mu$ is the one that extends $\mu_f$. You want to prove from scratch (without the Monotone Class Theorem or similar) that
$\mu (A)=\int _Af'd\lambda.\ $
Now, clearly $\nu $ defined by $\nu(A)=\int _Af'd\lambda\ $ is a positive measure since $f$ is increasing. Therefore
$A\subseteq B\Rightarrow \int _Af'd\lambda)\le \int _Af'd\lambda.$
Note that $\mu_f(I)=f(b)-f(a)\ $ for $any$ interval with endpoints $a,b$ and that the result is clearly true if $A=(a,b]\ $ or if $\mu(A)=\infty.$
If $A$ is Borel such that $\mu(A)<\infty,\ $ there is a sequence of disjoint intervals $\left \{ (a_i,b_i) \right \}_{i\in \mathbb N}$ such that $A\subseteq \bigcup_i(a_i,b_i)\ $ and
$\mu(A)>\mu \left ( \bigcup_i (a_i,b_i) \right )-\epsilon=\sum_i (f(b_i)-f(a_i))-\epsilon=\int _{\cup_i (a_i,b_i)}f'd\lambda-\epsilon>\int_Af'd\lambda-\epsilon,\ $ so
$\mu (A)\ge \int_Af'd\lambda.$
Similarly, there is a sequence of disjoint intervals $\left \{ (c_i,d_i) \right \}_{i\in \mathbb N}$ such that $A\subseteq \bigcup_i(c_i,d_i)\ $ and
$\nu(A)=\int_Af'd\lambda >\int _{\bigcup _i(c_i,d_i)}f'd\lambda -\epsilon=\sum_i(f(d_i)-f(c_i))-\epsilon =\mu (\bigcup _i(c_i,d_i))-\epsilon>\mu(A)-\epsilon,$ so
$\mu (A)\le \int_Af'd\lambda.$
The result follows.
Best Answer
First, your cover should consist of half-open intervals, not closed intervals. So what you mean is rather: $$\bigcup \limits_{k=1}^{\infty} [x,2^{-k}\cdot\varepsilon).$$
Then you may get a problem if $F$ is discontinuous at $x$. Remember it's only left continuous, so you could have $\delta>0$ such that $F(x+\epsilon)>F(x)+\delta$ for any $\epsilon>0$. Then $$\sum_{k=1}^{\infty} \lambda([x,2^{-k}\cdot\varepsilon))$$ would not converge.
Why don't you just pick $[x,x+\epsilon)$ for $k=1$ and $[x,x)$ for $k>1$. Then you don't get into that trouble.