Here is Theorem 7.32 in the book Mathematical Analysis - A Modern Approach to Advanced Calculus by Tom M. Apostol, 2nd edition:
Let $\alpha$ be of bounded variation on $[a, b]$ and assume that $f \in R(\alpha)$ on $[a, b]$. Define $F$ by the equation $$ F(x) = \int_a^x f \, d \alpha, \qquad \mbox{ if } x \in [a, b]. $$ Then we have:
i) $F$ is of bounded variation on $[a, b]$.
ii) Every point of continuity of $\alpha$ is also a point of continuity of $F$.
iii) If $\alpha \nearrow$ on $[a, b]$, the derivative $F^\prime(x)$ exists at each point $x$ in $(a, b)$ where $\alpha^\prime(x)$ exists and where $f$ is continuous. For such $x$, we have $$ F^\prime(x) = f(x) \alpha^\prime(x). $$
And, here is Apostol's proof of this theorem.
If suffices to assume that $\alpha \nearrow$ on $[a, b]$. If $x \neq y$, Theorem 7.30 implies that $$ F(y) - F(x) = \int_x^y f \, d \alpha = c \big[ \alpha(y) - \alpha(x) \big], $$ where $m \leq c \leq M$ (in the notation of Theorem 7.30). Statements (i) and (ii) follow at once from this equation. To prove (iii), we divide by $y-x$ and observe that $c \to f(x)$ as $y \to x$.
Finally, here is Theorem 7.30 (First Mean-Value Theorem for Riemann-Stieltjes Integrals):
Assume that $\alpha \nearrow$ and let $f \in R(\alpha)$ on $[a, b]$. Let $M$ and $m$ denote, respectively, the $\sup$ and $\inf$ of the set $\big\{ f(x) \colon x \in [a, b] \big\}$. Then there exists a real number $c$ satisfying $m \leq c \leq M$ such that $$ \int_a^b f(x) \, d \alpha(x) = c \int_a^b d \alpha(x) = c \big[ \alpha(b) - \alpha(a) \big]. $$ In particular, if $f$ is continuous on $[a, b]$, then $c = f \left( x_0 \right)$ for some $x_0$ in $[a, b]$.
Now my question is, although how (i) and (ii) follow is clear, how to gaurantee that $c \to f(x)$ as $y \to x$? After all, the only thing we know about $c$ from Theorem 7.30 in that $m \leq c \leq M$.
However, we can apply Theorem 7.30 to the interval with endpoints $x$ and $y$ and thus conclude that $c$ depends on $x$ and $y$ so that we then have $$ F(y) - F(x) = \int_x^y f \, d\alpha = c_{x, y} \big[ \alpha(y) - \alpha(x) \big]. $$
If $m = \inf \{ f(x) : x \in [a,b] \}$ and $M = \sup \{ f(x) : x \in [a,b] \}$, and $f$ is continuous, what would you conclude about $c$ as $y \to x$? Can you see why it is unnecessary to specify that $c$ is some function of $x$ and $y$?