= Solution
The stated bound is <Lipschitz continuity> with constant $1$. For every $x$ and every $\varepsilon>0$, taking $\delta=\varepsilon$ ensures $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$. The same $\delta$ works at every point, so $f$ is both continuous and <uniformly continuous>.
For an integer $N\ge1$, partition $[0,1]$ into $N$ equal intervals and define a <step function>
$$
g_N(x)=f(j/N)\quad\text{for }\frac jN\le x<\frac{j+1}{N},
\qquad j=0,\ldots,N-1.
$$
At the remaining endpoint set $g_N(1)=f((N-1)/N)$, so the final interval may be taken closed on the right. The <Lipschitz bound> gives
$$
|f(x)-g_N(x)|\le\frac1N
\quad\text{for all }x\in[0,1].
$$
Choosing $N\ge1/\varepsilon$ proves the requested <uniform step approximation on a compact interval>:
$$
\boxed{\sup_{x\in[0,1]}|f(x)-g_N(x)|\le\frac1N\le\varepsilon.}
$$
\b[A sufficiently fine partition gives a piecewise constant approximation uniformly over the whole interval.]
The <continuous function> $f$ and every finite <step function> $g_N$ are <Riemann integrable>. Put $c_j=f(j/N)$. Integrating each constant piece explicitly gives
$$
\int_0^1g_N(t)\cos(nt)\,dt
=\frac1n\sum_{j=0}^{N-1}c_j
\left[\sin\left(\frac{n(j+1)}N\right)
-\sin\left(\frac{nj}N\right)\right].
$$
Since the sine differences have absolute value at most $2$, the <triangle inequality> and uniform approximation give
$$
\begin{aligned}
|u_n|
&\le\left|\int_0^1(f-g_N)(t)\cos(nt)\,dt\right|
+\left|\int_0^1g_N(t)\cos(nt)\,dt\right|\\
&\le\frac1N+\frac2n\sum_{j=0}^{N-1}|c_j|.
\end{aligned}
$$
To prove convergence, fix $\varepsilon>0$ and first choose $N$ with $1/N<\varepsilon/2$. With this $N$ held fixed, the finite number $\sum_j|c_j|$ is independent of $n$, so the second term is also less than $\varepsilon/2$ for sufficiently large $n$. Therefore
$$
\boxed{u_n\longrightarrow0.}
$$
\b[First make the approximation error small, then let the oscillation frequency grow.] This is the <step-function proof of the Riemann-Lebesgue lemma>; it does not require assuming that a Lipschitz function has an everywhere-defined derivative.
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