Riemann-Stieltjes integral of a continuous function w.r.t. a step function

To suggest a better strategy and see the continuity of $f$ enter as it should, consider the proof for a single step discontinuity at $\alpha_1 \in (a,b)$. The objective is to show that for any $\epsilon > 0$ there exists $\delta > 0$ such that for any partition with $\|\Gamma\| < \delta$, we have $|R_\Gamma - f(\alpha_1)d_1| < \epsilon$.

Since $f$ is uniformly continuous on the compact interval $[a,b]$ , there exists $\delta > 0$ such that $|f(x) - f(y)| < \epsilon/|d_1|$ when $|x-y| < \delta$.

Take any partition $\Gamma$ with points $a = x_0 < x_1 < \ldots < x_n = b$ such that $\|\Gamma\|< \delta$. There exists and index j such that one of the following three cases must hold:

$$\tag{1} \quad x_{j-1} < \alpha_1 < x_j, \quad \varphi(x_{j-1}) = \varphi(\alpha_1^-),\quad\varphi(x_{j}) = \varphi(\alpha_1^+)$$ $$\tag{2} \quad x_{j-1} < \alpha_1 = x_j,\quad \varphi(x_{j-1}) = \varphi(\alpha_1^-),\quad\varphi(x_{j}) = \varphi(\alpha_1^+)$$ $$\tag{2} \quad x_{j-1} = \alpha_1 < x_j,\quad \varphi(x_{j-1}) = \varphi(\alpha_1^-),\quad\varphi(x_{j}) = \varphi(\alpha_1^+)$$

In each case we have for some intermediate point $\xi \in [x_{j-1},x_j]$,

$$\left| R_\Gamma - f(\alpha_1)d_1\right|= \left|f(\xi)(\varphi(x_j) - \varphi(x_{j-1})-f(\alpha_1) (\varphi(\alpha_1^+) - \varphi(\alpha_1^-) \right|\\ = |f(\xi) - f(\alpha_1)|\, |d_1|< \epsilon$$

To generalize the proof for multiple points where the step function is discontinuous, choose $\delta$ such that in addition to controlling the oscillation of $f$ on subintervals of partitions with $\|\Gamma\| < \delta$, we also have no more than one point of discontinuity $\alpha_k$ in a subinterval.

Actually this problem admits a direct solution. Notice that $$\int_a^b fd\phi=\sum_{i=1}^m\int_{\alpha_{i-1}}^{\alpha_i}fd\phi $$ because $\{\alpha_i\}$ it's a partition of $[a,b]$. Since $\phi$ is constant on every $[\alpha_{i-1},\alpha_i)$ and possible different only on $\alpha_i$, any Riemann Stieltjes sum is zero except for one single term (the right extreme), i.e., $$\int_{\alpha_{i-1}}^{\alpha_i}fd\phi=f(\alpha_{i})d_i $$ from where your result follows, here you need to use the uniform continuity of $f$. I realized that this idea is very similar to the other answer, with the advantage that if you prove the result for one single integral, then by additivity the general result follows. I will add the proof since it's a little bit different:

Let $\varepsilon>0$, and let $\delta>0$ such that the uniform continuity of $f$ holds on $[\alpha_{i-1}, \alpha_i]$ with $\varepsilon/|d_i|$ take $\Gamma=\{n_j\}_{j=0}^k$ a partition of $[\alpha_{i-1}, \alpha_i]$ with $\max_{1\leq j\leq k}\{|n_{j}-n_{j-1}|\}<\delta$ and $\{p_j\}$ a sequence of points such that $n_{j-1}\leq p_j\leq n_j$, then $$R_{\Gamma} =\sum_{j=1}^mf(p_j)(\phi(n_{j-1})-\phi(n_{j}))=f(p_k)(\phi(n_{k-1})-\phi(n_k))=f(p_k)(\phi(n_k^+)-\phi(n_k^-))$$ so we have $R_\Gamma=f(p_k)(\phi(\alpha_i^+)-\phi(\alpha_i^-))=f(p_k)d_i$ now since $n_{k-1}\leq p_k\leq \alpha_i$ we have $$|R_\Gamma-f(\alpha_i)d_i|\leq |d_i||f(p_k)-f(\alpha_i)|<|d_i|\varepsilon/|d_i|=\varepsilon $$ by the uniform continuity of $f$, this completes the proof.