This is a sketch presented in Apostol's Mathematical Analysis. Can you fill in the details? There aren't too many left! You need to look at the $S_1,S_2$ carefully.
Let $\int_a^b f(x)dx =I$, $M=\sup\{|f(x)|:x\in[a,b]\}$. Given $\epsilon >0$; choose $P_\epsilon$ such that $U(P_\epsilon,f)<I+\dfrac \epsilon 2$. [Here $U$ is the upper Darboux/Riemann sum of $f$] Let $N$ be the number of points of division in $P_\epsilon$ and let $\delta=\dfrac{\epsilon}{2MN}$. If $\Vert P\Vert <\delta$ put $$U(f,P)=\sum M_kf(x) \Delta_x=S_1+S_2$$ where $S_1$ is the sum over the terms that belong to the subintervals of $P$ that have no points of $P_\epsilon$ and $S_2$ is the remianing sum. Then $$S_1\leq U(P_\epsilon,f)<I+\frac \epsilon 2$$ $$S_2\leq NM\Vert P\Vert <NM\delta=\frac\epsilon 2$$ thus $$U(P,f)<I+\epsilon$$ Analogously, $$L(f,P)>I-\epsilon$$ if $\Vert P\Vert<\delta^\prime$, for a suitable $\delta^\prime$. Thus $$|S(P,f)-I|<\epsilon$$ if $\Vert P\Vert<\min\{\delta,\delta^{\prime}\}$.
NOTATION $U(f,P)$ denotes the upper sum of $f$ w.r.t $P$, that is $$\sum M_k \Delta x_k$$ where $M_i$ is the supremum of $f$ over the $i$-th subinterval of $P$. Analogously, $L(f,P)$ is the lower sum of $f$ w.r.t. $P$. Finally $S(f,P)$ is any Riemann sum of a tagged partition $P$.
That the first condition implies the first is immediate, since (using your notation) you always have $m_i \le f(\xi_i) \le M_i$, so the sums in the second definition are caught between the $L$ und $U$ sums.
Edit in response to a comment an additional explanation is necessary here. For this direction
it suffices to show that $I^* = lim_{||P||\rightarrow 0} L(f,P)$ and $I_* = lim_{||P||\rightarrow 0} U(f,P)$ Since both parts are similar it suffices to show, e.g., the first equality.
First it is easy to see that for partitions $P\subset P^\prime$ we have $L(f,P)\le L(f,P^\prime)$. A remaining hurdle is that for two partitions we do not necessarily know that one is a subset of the other one. This is resolved by looking at common refinements:
Assume $P$ satisfies $L(f,P) > I^* - \varepsilon$ and $Q$ is an arbitrary partition. We need to show that then there is a refinement $Q^\prime$ of $Q$ such that $L(f,Q^\prime)\ge L(f,P)$ (and, consequently, $L(f,Q^\prime)>I^*-\varepsilon$).
For $Q^\prime$ one can choose the common refinement $R$: if $P=\{x_1,\ldots x_n \}$ and $Q=\{y_1,\ldots y_m \}$ then we just let $R = P\cup Q$. Since this is a refinement of both $P$ and $Q$ we have both $L(f,R)\ge L(f,P)$ as well as $L(f,R)\ge L(f,Q)$
Second edit: the original version was not correct:
For the other direction it suffices to show that if the function is integrable in the sense of the second definition then both $I_*$ and $I^*$ agree with the of the sums from the second definition. Since the reasoning is the same in both cases I'll just look at $I_*$.
So fix $\varepsilon >0$ and a given partition $P$ such that
$$|L - \sum_{i=1}^n f(\xi_i)\Delta x_i |< \varepsilon$$
if only the partition is fine enough.
Choose such a partition $P=\{x_0,\dots x_n\}$ and to $[x_{i-1},x_{i}]$ choose $\eta_i\in[x_{i-i},x_{i}]$ such that for
$m_i:=\inf \{ f(x):x\in [x_{i-1},x_i]\} $
we have $$0\le f(\eta_i)-m_i\le \frac{\varepsilon}{2n}$$
Then
\begin{eqnarray}
| L -\sum_{i=1}^n m_i \Delta x_i|
& = & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i + \sum_{i=1}^n f(\eta_i)\Delta x_i
-\sum_{i=1}^n m_i\Delta x_i| \\
&\le & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i| + \sum_{i=1}^n | f(\eta_i)
- m_i|\Delta x_i \\
& < & \frac{\varepsilon}{2} + \sum_{i=1}^n \frac{\varepsilon}{2n}=\varepsilon
\end{eqnarray}
If you 'see' that $0 <L -I_*< L -\sum_{i}m_i \Delta x_i$ then you are done here, otherwise it follows easily from the last estimate that the $\sum_i m_i \Delta x_i$ are, for any partition which is fine enough, $\varepsilon $ close to the fixed real number $L$, which of course implies that the $\sup$ over these sums exists and equals $L$ (here you need to use again the fact that you will approach the $\sup$, if it exists, if the width of the partitions goes to $0$).
Best Answer
It is the limit of a net. Nets are a generalization of sequences which make all the familiar statements about sequences true for spaces that are not first-countable (for example a point lies in the closure of a subspace if and only if there is a net converging to it, and so forth), so any time you want to prove something about general spaces and you would like to use sequences but can't, you can use nets instead (although there are some subtleties here; one cannot just replace "sequence" with "net" in a proof).