Prove by contradiction. Suppose that there exists $x_{0}\in[a,b]$
such that $f$ is continuous at $x_{0}$ and $f(x_{0})>0$. Let $c=\frac{1}{2}f(x_{0})>0$.
Since $f$ is continuous at $x_{0}$ and $f(x_{0})>c$, there exists
$\delta>0$ such that $f(x)>c$ for any $x\in[x_{0}-\delta,x_{0}+\delta]$.
(We need to adjust the interval $[x_{0}-\delta,x_{0}+\delta]$ accordingly
if $x_{0}=a$ or $x_{0}=b$.)
Since $f\geq0$, we have
$$
\int_{a}^{b}f(x)\,dx\geq\int_{x_{0}-\delta}^{x_{0}+\delta}f(x)\,dx\geq\int_{x_{0}-\delta}^{x_{0}+\delta}c\,dx=2\delta c>0,
$$
which is a contradiction.
No, it is not correct.
First of all, you seem to think that $\inf_{[-1,1]}f=-1$. Actually, $\inf_{[-1,1]}f=0$.
After writing that$$\sup_{[-1,1]}f-\inf_{[-1,1]}f=2,\tag1$$you use no other property of the function in your proof. Therefore, if your proof was correct, it would prove that any function for which $(1)$ holds is Riemann-integrable. That's not the case. If, say,$$f(x)=\begin{cases}1&\text{ if }x\in\Bbb Q\\-1&\text{ otherwise,}\end{cases}$$then $f$ is not Riemann-integrable, although $(1)$ holds.
Your error lies in assuming that, for every $\varepsilon>0$ and every $n\in\Bbb N$, there is a partition $P$ of $[-1,1]$ into $n$ intervals with $|P|<\frac\varepsilon{2n}$.
Your function is actually Riemann-integrable. To see why, take $\varepsilon>0$. Since $\frac\varepsilon2>0$ and since $f(x)=1$ only for finitely many points of $\left[\frac\varepsilon2,1\right]$, it's not hard to see that there is a partition $P=\left\{a_0\left(=\frac\varepsilon2\right),a_1,\ldots,a_n(=1)\right\}$ of $\left[\frac\varepsilon2,1\right]$ such that$$\overline\sum\left(f_{\left[\frac\varepsilon2,1\right]},P\right)-\underline\sum\left(f_{\left[\frac\varepsilon2,1\right]},P\right)<\frac\varepsilon2.$$But then, if $P^\star=\left\{-1,0,a_0,a_1,\ldots,a_n\right\}$, then$$\overline\sum\left(f_{\left[-1,1\right]},P^\star\right)-\underline\sum\left(f_{\left[-1,1\right]},P^\star\right)<\varepsilon.$$
Best Answer
Your idea is good. For given $\varepsilon > 0$ we have to find $\delta > 0$ such that for all $0 < c < \delta$ we have $\left|\int_c^1 f(x) \mathrm{d}x - \int_0^1 f(x) \mathrm{d}x\right| < \varepsilon$. Take the partition $P_\varepsilon$ of $[0,1]$ and $\delta$ as you suggested. For any $0 < c < \delta$, $P_\varepsilon$ gives rise to a partition $P_\varepsilon'$ of $[c,1]$ of smaller or equal mesh size by replacing $x_0 = 0$ by $x_0' = c$. Refine $P_\varepsilon'$ in such a way that for the new partition $P' \supseteq P_\varepsilon'$ we have that all corresponding Riemann sums approximate the integral up to an error of $\varepsilon$. $P = P \cup \lbrace x_{-1} := 0 \rbrace \supseteq P_\varepsilon$ is partition of $[0,1]$. Riemann sums corresponding to $P$ and $P'$ differ by the term $f(t_{-1})(x_0 - x_{i-1})$, whose absolute value is bounded by $\|f\|_\infty \delta$.
An easier way requires a little more knowledge about Riemann integrals, but is considerably shorter: We have $$\left|\int_c^1 f(x) \mathrm{d}x - \int_0^1 f(x) \mathrm{d}x\right| = \left|\int_0^c f(x) \mathrm{d}x\right| \leq \|f\|_\infty c \to 0, \quad c \to 0.$$