Proof
(related to Lemma: Approximability of Continuous Real Functions On Closed Intervals By Step Functions)
 Let $[a,b]$ be a closed real interval.
 Let $\mathbb R$ be the set of real numbers.
 Let \(f:[a,b]\mapsto \mathbb R\) a continuous function.
 It is to be shown that for every \(\epsilon > 0\) there exist some step functions \(\phi:[a,b]\mapsto\mathbb R\) and \(\psi:[a,b]\mapsto\mathbb R\) with $(i)$ $\phi(x) \le f(x)\le \psi(x)$ and $(ii)$ $\phi(x)\psi(x)\le\epsilon$ for all $x\in[a,b].$ This will be shown by construction.
 Because all continuous real functions on closed intervals are uniformly continuous, so is $f$.
 This means that for every $\epsilon > 0$ there is a $\delta > 0$ such that $f(x)f(y) < \frac\epsilon2$ for all $x,y\in D$ with $xy < \delta.$
 Because of the existence of arbitrarily small positive rational numbers, we can choose a natural number \(n\) big enough such that $\frac 1n < \delta/(ba)$, or, equivalently $\frac {ba}n < \delta$.
 By setting $$t_k:=a+k\frac{ba}n,\quad k=0,1,\ldots, n,$$ we get an equidistant partition of the closed interval $[a,b]$ with $a=t_0 < t_1 < \ldots < t_n=b.$
 Set $\phi(a):=\psi(a):=f(a)$, and for $k=1,\ldots,n$ and $t_{k1} < x\le t_k$ set $$\phi(x):=f(t_k)\frac\epsilon2,\quad \psi(x):=f(t_k)+\frac\epsilon2.$$
 By construction, it follows that $\phi(x)\psi(x)\le\epsilon$ for all $x\in[a,b],$ thus $(2)$ is proven.
 For all \(x\in ]t_{k1},t_k]\) we have by construction $$\frac \epsilon2 < f(x)f(t_k) < \frac \epsilon2,$$ thus $$\phi(x) = f(t_k)\frac \epsilon2 < f(x) < f(t_k) + \frac \epsilon2=\psi(x),$$ therefore $\phi(x) \le f(x)\le \psi(x)$ for all $x\in[a,b],$ thus $(1)$ is proven.
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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983