We can use the following:
Lemma. If $f$ is continuous on $[a,b]$ then for any $\epsilon > 0$ there exists $\delta > 0$ such that for any partition $P$ of $[a,b]$ with $\|P\| < \delta$ and any refinement $R$ of $P$, we have $|S(f,P) - S(f,R)| < \epsilon$.
Applying the lemma, we can show that if $f$ is continuous, then the Cauchy criterion is satisfied. That is, for any $\epsilon >0$ there exists $\delta > 0$ such that if $P$ and $Q$ are any partitions satisfying $\|P\|, \|Q\| < \delta$, then $|S(f,P) - S(f,Q)| < \epsilon.$
To see this, let $R = P \cup Q$ be a common refinement and take $\delta$ as specified in the lemma such that if $\|P\|, \|Q\| < \delta$, we have $|S(f,P) - S(f,R)| < \epsilon/2$ and $|S(f,Q) - S(f,R)| < \epsilon/2$. Whence, it follows that
$$|S(f,P) - S(f,Q)| \leqslant |S(f,P) - S(f,R)| + |S(f,Q) - S(f,R)| < \epsilon/2 + \epsilon/2 = \epsilon.$$
It remains to prove the lemma.
Since $[a,b]$ is compact, $f$ is uniformly continuous and for any $\epsilon > 0$ there exists $\delta >0$ such that if $|x - y| < \delta$, then $|f(x) - f(y)| < \epsilon/(b-a)$. Suppose $\|P\| < \delta$ and $R$ is a refinement of $P$. Any subinterval $[x_{j-1}, x_{j}]$ of $P$ can be decomposed as the union of subintervals of $R$,
$$[x_{j-1},x_{j}] = \bigcup_{k=1}^{n_j}[y_{j,k-1}, y_{j,k}],$$
and
$$\begin{align}\left|S(f,P) - S(f,R)\right| &= \left|\sum_{j=1}^n f(\xi_j)(x_j - x_{j-1}) - \sum_{j=1}^n \sum_{k=1}^{n_j}f(\eta_{j,k})(y_{j,k} - y_{j,k-1})\right| \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}|f(\xi_j) - f(\eta_{j,k})|(y_{j,k} - y_{j,k-1}) \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}\frac{\epsilon}{b-a}(y_{j,k} - y_{j,k-1}) \\ &= \epsilon\end{align}$$
Hint
For $0 \lt \epsilon \lt 1$, consider the subdivision
$$\begin{aligned}\sigma \equiv x_ 0 &= 0 \lt x_1 = \frac{\epsilon}{2}\\&\lt x_2 = \frac{1}{N} -\frac{\epsilon}{2 2^{N+1}} \lt x_3 = \frac{1}{N} +\frac{\epsilon}{2 2^{N+1}}\\
&\dots\\
&\lt x_{N} = \frac{1}{2} -\frac{\epsilon}{2 2^{2}} \lt x_3 = \frac{1}{N} +\frac{\epsilon}{2 2^{2}}\\
&\lt x_{N+2} = \frac{1}{2^0} -\frac{\epsilon}{2 2^{1}} \lt x_{N+3} = 1
\end{aligned}$$
Where $N$ is the largest integer such that $\frac{1}{N}\gt \epsilon / 2$.
To understand why that works, I suggest that you take an example, i.e. $\epsilon = 1/10$, and make a drawing of the subdivision vs. the function. From there, you can compute the Darboux lower and upper sum to complete the proof. You'll essentially see that Darboux lower sum is equal to zero while Darboux upper sum is less than
$$\frac{\epsilon}{2} + \frac{\epsilon}{2}\left(\frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{N+1}}\right) \lt \epsilon.$$
The idea, as you have seen is to isolate zero on one side from the other discontinuities of the function.
Best Answer
The computations are correct (although I think that it would be more interesting to do it without computing the Riemann integral of another function). However, there is a problem when you state that$$\int_{-2}^3f(x)\,\mathrm dx=\int_{-2}^0f(x)\,\mathrm dx+\int_0^3f(x)\,\mathrm dx.$$Your goal is to prove that $f$ is not integrable, but here you are acting as if it was. You could simply clam that if $f$ is not Riemann integrable on a subinterval of $[-2,3]$, then it is also not Riemann integrable on $[-2,3]$.