Your definition is
$(1)$ $f\in \mathscr R[a,b]$ with integral $\int_a^b f$ if and only if for each $\epsilon >0$ there exists $\delta >0$ such that for any partition $P$ with $\lVert P\rVert <\delta$ we have that $$\left|\int_a^b f-\sum(f,P)\right|<\epsilon$$
where $\sum(f,P)$ means the Riemann sum of $f$ with respect to $P$. The alternative definition is
$(2)$ $f\in \mathscr R[a,b]$ with integral $\int_a^b f$ if and only if for each $\epsilon >0$ there exists a partition $P_\epsilon$ such that for any partition $P$ a refinement of $P_\epsilon$ we have that $$\left|\int_a^b f-\sum(f,P)\right|<\epsilon$$
We would like to show one implies the other. First
$(1) \implies (2)$ Suppose $(1)$ holds. Since refinements can only decrease the mesh, and $\delta$ depends on $\epsilon$, the claim follows: for each $\epsilon>0$ take a partition $P_\epsilon$ with mesh $\delta'<\delta$ given by the above. Then any refiniment will have mesh at most $\delta'$ which will be less than $\delta$, and hence $(2)$ will hold.
$(2)\implies (1)$ This one is the tricky one. Here you can find a proof
An easy characterization of Riemann integrability is the following, which doesn't requiere that we know what the value of integral is:
$f:[a,b]\to\Bbb R$ is Riemann integrable over $[a,b]$ if and only if for each $\epsilon >0$ there exist a partition $P_\epsilon$ such that for any refinement $P$ of $P_\epsilon$ $$U(f,P)-L(f,P)<\epsilon$$
Moreover, this is equivalent to $$\overline{\int_a^b} f=\underline{\int_a^b}f$$
so if you're able to prove the above and evaluate any upper or lower integral, you're done.
In fact, this applies to the Riemann Stieljes integral whenever the integrator $\alpha$ is montone.
Let $\alpha:[a,b]\to\Bbb R$ be monotone. Let $f:[a,b]\to\Bbb R$. Then the following are equivalent:
$(1)$ The function $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ over $[a,b]$
$(2)$ For each $\epsilon >0$ there exists a partition $P_\epsilon$ such that for any refinement $P$ of $P_\epsilon$ $$U(f,\alpha,P)-L(f,\alpha,P)<\epsilon$$
$(3)$ $$\overline{\int_a^b} f=\underline{\int_a^b}f$$
Hint. To show that $|f|$ is integrable Riemann, it suffices to show that for every partition $P$ of $[a,b]$
$$
U(|f|,P)-L(|f|,P)\le U(f,P)-L(f,P),
$$
which in order to prove it suffices to show that, for every $[s,t]\subset [a,b]$
$$
\sup_{x,y\in[c,d]} |f(x)|-|f(y)|=
\sup_{x\in[c,d]}|f(x)|- \inf_{x\in[c,d]}|f(x)|\le
\sup_{x\in[c,d]}f(x)- \inf_{x\in[c,d]}f(x)=\sup_{x,y\in[c,d]} f(x)-f(y),
$$
which is a consequence of the fact that
$$
\big| |f(x)|-|f(y)|\big|\le |f(x)-f(y)|.
$$
For the second part of the question, just use the fact that
$$
-|f(x)|\le f(x)\le |f(x)|
$$
and hence
$$
\int |f| \le \int f \le \int |f|
$$
and finally
$$
\Big|\int f\,\Big| \le \int|f|
$$
Best Answer
Let $R = c (b-a)$ and let $\varepsilon > 0$.
You want to show that there is $\delta > 0$ such that for all tagged partitions $P$ with $\|P\| < \delta$ you have $$ \left | \sum_{k=1}^n f(x_i) (t_{i+1}-t_i) - R \right | < \varepsilon$$
where $x_i \in [t_i , t_{i+1}] \subset [a,b]$ form a tagged partition of $[a,b]$. We have $f(x_i) = c$ hence $$ \left | \sum_{k=1}^n f(x_i) (t_{i+1}-t_i) - R \right | = \left |c \sum_{k=1}^n (t_{i+1}-t_i) - R \right | = \left | c (b-a) - R \right | = 0 < \varepsilon$$
hence $f$ is Riemann integrable with $$ \int_a^b f(x) dx = c (b - a)$$
Hope this helps.