In other words, how to prove
A continuous function over a closed interval is Riemann-integrable. That is, if a function $f$ is continuous on an interval $[a, b]$, then its definite integral over $[a, b]$ exists.
Edit:
The Definite Integral as a Limit of Riemann Sums :
Let $f(x)$ be a function defined on a closed interval $[a, b]$. We say that a number $I$ is the definite integral of $f$ over $[a, b]$ and that $I$ is the limit of the Riemann sums $\sum \limits_{k=1}^n f(c_k)\Delta x_k$ if the following condition is satisfied:
Given any number $\epsilon \gt 0$, there is a corresponding number $\delta \gt 0$ such that for every partition $P = \{x_0, x_1, … , x_n\}$ of $[a, b]$ with $\|P \| < \delta$ and any choice of $c_k$ in $[x_{k-1}, x_k]$, we have
$$ \left| \sum_{k=1}^n f(c_k) \Delta x_k – I \ \right| \lt \epsilon .$$
Best Answer
First note that a precise formulation of your question is:
You can find a proof in Chapter 8 of these notes.
Here is a rough outline of this handout:
I. I introduce the ("definite") integral axiomatically. One of the axioms is that the set of integrable functions on $[a,b]$ should contain all the continuous functions.
II. I prove that the Fundamental Theorem of Calculus follows (easily) from the axioms.
III. I introduce Riemann integrable functions (which are exactly what you wrote above) and verify that the class of Riemann integrable functions on $[a,b]$ satisfies the axioms of I. In particular:
IV: I prove that every continuous function is Riemann integrable.
Later I talk about the Darboux integral and how it compares to the Riemann integral. But it was an intentional decision to present the Riemann integral first. This is what students are expecting from their previous courses, and it is not so bad to work with, at least for a while.