Here's what I'm trying to prove.
Let $f: [a,b] \to \mathbb{R}$ be a bounded function. Let $P$ and $Q$ be partitions of $[a,b]$ such that $P \subseteq Q$ ($Q$ is a refinement of $P$). Then:
$$L(f,P) \leq L(f,Q) \leq U(f,Q) \leq U(f,P)$$
I'm only going to prove the result for lower sums just because I think a similar argument can be made for upper sums.
Proof Attempt:
Let $|P| = n$ and let $|Q| = n+m$ for some $m \in \mathbb{N}_0$. We proceed by induction on $m$.
If $m = 0$, then $P = Q$ so $L(f,P) = L(f,Q)$. That proves the result in the base case.
Now, suppose that it holds for arbitrary $m$. Consider $Q'$ so that $Q \subseteq Q'$ and $|Q'| = n+(m+1)$. So, we have picked a specific point $x' \in [a,b]$ to be included in $Q'$. That just means that $Q' = Q \cup \{x'\}$. Then:
$$\exists i \in \{1,2,\ldots,n+m\}: x' \in [x_{i-1},x_i]$$
Let $A = \inf \{f(x): x \in [x_{i-1},x'] \}$ and $B = \inf \{f(x): x \in [x',x_{i-1}]$. Let $m_i = \inf\{f(x): x \in [x_{i-1},x_i] \}$. Then:
$$A \geq m_i \land B \geq m_i$$
$$A(x'-x_{i-1}) \geq m_i (x'-x_{i-1}) \land B(x_i-x') \geq m_i (x_i-x')$$
$$A(x'-x_{i-1}) + B(x_i-x') \geq m_i (x_i-x')$$
$$A(x'-x_{i-1}) + B(x_i-x') - m_i (x_i-x') \geq 0$$
We notice that the other terms in the sum $L(f,Q)$ were unchanged and so, we get the following:
$$L(f,Q')-L(f,Q) \geq 0$$
$$L(f,Q') \geq L(f,Q)$$
which implies that $L(f,Q') \geq L(f,P)$. So, the inequality holds for the case $m+1$. Hence, the given statement holds for all $m \in \mathbb{N}_0$. $\Box$
Does the proof above work? If it doesn't, then why? How can I fix it?