Suppose that $f:[a,b]$ is a continuous function. Since the interval is compact, $f$ is uniformly continuous, which means that for every $\varepsilon \gt 0 $ a $\delta > 0$ exists, so that for all $x,y \in [a,b]$ with $\lvert x-y\rvert \lt \delta$ applies: $\lvert f(x)-f(y)\rvert \lt \varepsilon$
Now I want to prove that a step function $S: [a,b] \to \mathbb{R}$ exists, so that $\lvert S(x) -f(x)\rvert \lt \varepsilon$ for every $x \in \mathbb{R}$ . Can you give me some starting-assist?
Split $[a,b]$ into small enough intervals, $[a, a+\delta), [a+\delta, a+2\delta),\dots$, then set $S(x)$ be equal to $f(\bar x)$ where $\bar x$ is the middle point of the interval in which $x$ lies.
Then use uniform continuity and triangular inequality to show that $|S(x) - f(x)|$ is small.