I wanted to provide a proof for the claim in the above. Namely, take a sequence of real valued functions $\{f_n\}_{n \in \mathbb{N}}$ converging uniformly to function $f$. These functions are all defined on common closed interval of the real line $[a,b]$. Further, assume each function in the sequence is Riemann-Stieltjes integrable with respect to increasing function $\alpha: [a,b] \rightarrow \mathbb{R}$.
Here is my proof that $f$ is also Riemann-Stieltjes integrable with respect to $\alpha$.
Choose any $\epsilon > 0$. As $f_n$ converges uniformly to $f$, there exists $N \in \mathbb{N}$ such that $n \geq N$ implies $|f_n(x) - f(x)| < \epsilon$ for all $x \in [a,b]$. Also, we know that there exists a partition of $[a,b]$, $P = \{x_1, x_2, ..., x_K\}$, for which $U(f_N, P, \alpha) - L(f_N, P, \alpha) < \epsilon$.
We get the inequality
\begin{align}U(f, P, \alpha) - L(f, P, \alpha) &= \sum_{i=1}^{K-1} (\sup_{[x_i, x_{i+1}]} f - \inf_{[x_i, x_{i+1}]} f )(\alpha(x_{i+1}) - \alpha(x_i))\\ &< \sum_{i=1}^{K-1} (\sup_{[x_i, x_{i+1}]} f_N - \inf_{[x_i, x_{i+1}]} f_N + 2\epsilon )(\alpha(x_{i+1}) - \alpha(x_i))\\ &< \epsilon + 2\epsilon(\alpha(b) - \alpha(a))\\ &= \epsilon(2(\alpha(b) - \alpha(a)) + 1) \end{align}
which proves the claim. Can someone let me know if this proof is valid?