For this, I suppose that $f:[a,b]\to \mathbb{R}$, and $P=\{a=x_0\lt x_1\lt\ldots\lt x_m=b\}$ is a partition of $[a,b]$. Take care of the considerations about partitions given by Dyland's comment. Let $S_m=\{1,2,\ldots, m\}.$ Note that any partition $P^*$ finer than $P$ can be obtained from $P$ by adjoining it some points, we say $n$. So, we proceed by induction on $n$.
Suppose that $P^*$ is obtained by adding a point $x$ to $P$. Then $x\in (x_{r-1},x_r)$, for some $r\in S_m$. Let $m_i=\inf f([x_{i-1},x_i]) \text{ for } i\in S_m$ m_r'=\inf f([x_{r-1},x]) \text{ and } m_r''=\inf f([x,x_r]) Then \begin{align*} L(f,P^*)-L(f,P) &= m_r'(x-x_{r-1}) + m_r''(x_r-x)-m_r(x_r-x_{r-1})\\ &= (m_r'-m_r)(x-x_{r-1}) + (m_r''-m_r)(x_r-x)\\ &\geq 0. \end{align*} You can verify that the factors that involve infs are nonnegative, and this proves the claim. I'll leave the inductive step to you.
For upper sums is similar. For the Riemann-Stieltjes integral the lower and upper sums are studied usually for increasing integrators $\alpha$, and then the proof is the same thing.