Proof that $\forall aI don't understand a lot of parts of the proof where. In blockquotes will the be proof provided by my professor, followed by my questions.

Let $\epsilon >0$ and let $N > \frac{(b-a)^2}{\epsilon}$. Then let $P_N = \{ a + \frac{0 (b-a)}{N} + .... a + \frac{ N (b-a)}{N} \}$.

Why do we select this partition? I can see why this works later on, but what is the intuition? 

$U(f,p) = \sum (a + \frac{i (b-a)}{N} \cdot \frac{b-a}{N}
= Na \cdot \frac{b-a}{N} + \frac{(b-a)^2}{N^2} \sum_{i=1}^N i $

I understand this is the definition of $U(f,p)$ and we get $Na$ from the sum.

$(a(b-a) + \frac{(b-a)^2}{N^2} \cdot \frac{(N)(N+1)}{2}$

Okay, this is where I get lost. Where did this term $\frac{(N)(N+1)}{2}$ come from? I'm guessing it must have something to do with the sum but I (for the life of me) cannot figure it out. 

$=a(b-a) + \frac{(b-a)^2}{2} + \frac{(b-a)^2}{2N}
>\frac{1}{2} (b^2-a^2) + \frac{(b-a)^2}{2N} < \frac{1}{2} (b^2-a^2) + \epsilon$

I'm also lost at the first line above. As for the second line, I understand we are plugging our value for $N$ in order to get an $\epsilon$. 
 A: Note that on $[c,d]$, we have $c \le f(x) \le d$ and so $(\sup_{x \in [c,d]} f(x)) (d-c) = d (d-c)$ and $(\inf_{x \in [c,d]} f(x)) (d-c) = c (d-c)$.
Note that $1+2+\cdots + m = {1 \over 2} m (m+1)$.
Now consider the partition $P_n$ with the points $a+{k \over n} (b-a)$, then we have
\begin{eqnarray}
L(f,P_n) &=& \sum_{k=0}^{n-1} (a + {k \over n} (b-a) ) {1 \over n} (b-a) \\
&=& {1 \over n}(b-a) [ an + ({b-a \over n}) {1 \over 2} n(n-1)] \\
&=& (b-a) [ a + {b-a \over 2} (1-{1 \over n})]
\end{eqnarray}
and we see that $\lim_n L(f,P_n) = {1 \over 2}(b^2-a^2)$.
A similar analysis shows that $\lim_n U(f,P_n) = {1 \over 2}(b^2-a^2)$.
Since $L(f,P_n) \le \sup_{\cal P} U(f,P_n) \le \inf_{\cal P} U(f,P_n) \le U(f,P_n) $, we see that
$\sup_{\cal P} U(f,P_n) =  \inf_{\cal P} U(f,P_n) = {1 \over 2}(b^2-a^2)$
and so $f$ is integrable with integral ${1 \over 2}(b^2-a^2)$.
A: In fact we select the uniform subdivision $x_i=a+i\space(\frac{b-a}{n})$ because we can calculate the Riemann sum without too much difficulties.
There are also other choices that leads to convenient calculacy, for instance $x_i=a\space(\frac{b}{a})^{i/n}$ when $a>0$, because it leads to summing geometric series.
Anyway with a generic subdivision $a=x_0<x_1<...<x_n=b$, in this particular case you can cheat the calculation with this trick :
$S(f)=\sum_{i=1}^{n}(x_i-x_{i-1})f(t_i)$ for $t_i\in[x_{i-1},x_i]$.
Notice that $x_{i-1}\le f(t_i)\le x_i$.
So we have the usual rectangle sums:
the lower one $L(f)=\sum_{i=1}^{n}(x_i-x_{i-1})x_{i-1}$ 
the higher one $H(f)=\sum_{i=1}^{n}(x_i-x_{i-1})x_i$. 
We have $L(f)\le S(f)\le H(f)$ and both rectangle sums converge to $S(f)$.
Thus if $f$ is Riemann integrable then $\frac{H(f)+L(f)}{2}\to S(f)$ when the subdivision is fine enough.
It turns out that this is easy to calculate, because terms are nullifying each other, except two:
$\frac{H(f)+L(f)}{2} = \frac{1}{2}\sum_{i=1}^{n}(x_ix_{i-1}-x_{i-1}^2)+(x_i^2-x_ix_{i-1})=\frac{1}{2}\sum_{i=1}^{n}(x_i^2-x_{i-1}^2)=$
$\frac{1}{2}(x_n^2-x_0^2) = \frac{1}{2}(b^2-a^2)$
