Show that $\int_{[-1, 1]} f d sgn = 2 f(0).$ (Analysis 1 by Tao)

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Exercise 11.8.5. Let $sgn : \mathbb{R} \to \mathbb{R}$ be the signum function; $sgn(x) = 1$ when $x >0$, $=0$ when $x=0$, and $=-1$ when $x<0$. Let $f : [-1, 1] \to \mathbb{R}$ be a continuous function. Show that $f $is Riemann-Stieltjes integrable with respect to $sgn$, and that $$\int_{[-1, 1]} f d sgn = 2 f(0).$$ (Hint: for every $\epsilon >0$, find piecewise constant functions majorizing and minorizing $f$ whose Riemann-Stieltjes integral is $\epsilon$-close to $2f(0)$. )

Attept: Since $f$ is continuous on a closed interval, it is uniformly continuous. From the previous exercise, we know that $f$ is Riemann-Stieltjes integral. For the next part, since $f$ is bounded, $f([-1, 1]) \subset [-M, M]$ for some $M \in \mathbb{R}^+$. Therefore, let $h(x) = -M$ and $g(x) = M$ for all $x$. Then, $h$ and $g$ minorizes and majorizes $f$, but I don't think that this helps to show $f$ is $\epsilon$-close to $2f(0)$.

I appreciate if you give some help.