Prove $U(f,P')\le U(f,P)$ ($P'$ a refinement of $P$) with weird HINT In Calculus, there is a theorem that a refinement of a partition will produces a smaller upper sum. The proof is easy, and I know it. However, I found a handout from my teacher, see the figure.

In (b), it seems a much more simpler version of the theorem I mentioned above, since the refinement only added one point. However, I totally can't get how to prove (b) "from (a)". The hint said "Consequently," which means it is a simple inference directly from the theorem in (a). (Right?) But how can one use (a) to prove (b)? I have been thinking it for 30 minutes, but still being confused.
 A: You just need to consider the Riemann sum of $f$ over partition $P''=\{x_i, y, x_{i+1}\} $ of $[x_i, x_{i+1}]$. By $(a) $ we have $$U(f, P'')\leq M_{i+1}(x_{i+1}-x_i)=f(\xi_{i+1})(x_{i+1}-x_i) $$ and adding terms corresponding to intervals $$[x_0,x_1],\dots,[x_{i-1},x_i],[x_{i+1},x_{i+2}],\dots, [x_{n-1},x_n]$$ to the above inequality we get $U(f, P') \leq U(f, P) $.
I have deliberately stayed away from the notation used in your image as it is not standard. 
A: Observe that if $P:=\{ \xi_i \}_{i=0}^n$ is a partition of $[a,b]$, and $\xi_j= \zeta_0 < \zeta_1<...< \zeta_{l_j}=\xi_{j+1}$, then by (a) you know that:
$\underset{x\in [\xi_j,\xi_{j+1}]}{\sup} f(x)\cdot (\xi_{j+1}-\xi_{j})\geq \underset{r=1}{\overset{l_j}{\sum}} \Bigg( \underset{x\in [\zeta^j_{r},\zeta^j_{r+1}]}{\sup} f(x)\cdot (\zeta^j_{r+1}-\zeta^j_{r})\Bigg)$ 
Considering $P'$ which is a refinement of $P$ given by $\xi_j=\zeta_0^j<....<\zeta_{l_j}^j=\xi_{j+1}$ for all $0\leq j\leq n-1$, then:
$U(f,P) = \underset{j=1}{\overset{n}{\sum}} \Bigg( \underset{x\in [\xi_j,\xi_{j+1}]}{\sup} f(x)\cdot (\xi_j-\xi_{j+1})\Bigg)\leq$
$\leq \underset{j=1}{\overset{n}{\sum}} \Bigg( \underset{r=1}{\overset{l_j}{\sum}} \Big( \underset{x\in [\zeta^j_{r},\zeta^j_{r+1}]}{\sup} f(x)\cdot (\zeta^j_{r+1}-\zeta^j_{r})\Big) \Bigg)=U(f,P')$
