The definition of upper and lower integrals is $\mathcal{U}(f) = \inf_{P'} U(f,P')$ and $\mathcal{L}(f) = \sup_{P'} L(f,P')$ where the infimum and supremum are taken over all partitions $P'$ of $[a,b]$.
In other words, the upper integral $\mathcal{U}(f)$ is the greatest lower bound of all upper sums and for any particular partition $P$ we have $U(f,P) \geqslant \mathcal{U}(f)$. Similarly we have $L(f,P) \leqslant \mathcal{L}(f)$.
Thus,
$$\tag{*}U(f,P) - L(f,P) \geqslant \mathcal{U}(f) - \mathcal{L}(f)$$
In proving the reverse implication you first assume that $f$ is not Riemann integrable. This implies that $\mathcal{U}(f) > \mathcal{L}(f)$ and so $\alpha := \mathcal{U}(f) - \mathcal{L}(f) > 0$.
Take $\epsilon = \alpha/2$. By (*) it follows that for any partition $P$ we have
$$U(f,P) - L(f,P) \geqslant \mathcal{U}(f) - \mathcal{L}(f) = \alpha > \frac{\alpha}{2} = \epsilon$$
This contradicts the hypothesis that for any $\epsilon$ there is a partition $P$ such that $U(f,P) - L(f,P) < \epsilon$. Therefore, $f$ must be Riemann integrable.
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
Since $\int_{a}^{b}f = inf\{U(f,P)\}$, we can keep finding partitions that such that $U(f,P)$ are closer and closer to $\int_{a}^{b}f$ approaching from above. So choose a partition, call it $P_1$, such that it is closer than $\frac{\epsilon}{2}$ to $\int_{a}^{b}f$. That means $$ U(f,P_1) < \int_{a}^{b}f +\frac{\epsilon}{2}$$ $$ \iff U(f,P_1) - \int_{a}^{b}f <\frac{\epsilon}{2}$$
Finally note that $U(f,P_1) \geq inf\{U(f,P)\}$, so $$ 0 = \int_{a}^{b}f - \int_{a}^{b}f = inf\{U(f,P)\} - \int_{a}^{b}f \leq U(f,P_1) - \int_{a}^{b}f <\frac{\epsilon}{2}$$ $$ \iff 0 \leq U(f,P_1) - \int_{a}^{b}f <\frac{\epsilon}{2} $$
The argument is almost identical for $L(f,P)$ (except flipping some inequalities)