Consider $f$ be a continuous function on $[0,1]$. Let $\{f_{n}\}$ be a sequence of functions defined by $$f_n (x) = \begin{cases} f(\zeta_k),& x\in[\frac{k}{n},\frac{k+1}{n}),\\ 0, &\text{otherwise}.\end{cases}$$ for some $\zeta_{k}\in (\frac{k}{n},\frac{k+1}{n})$.
I want to prove that $f_n \rightarrow f$ uniformly on $[0,1]$.
Since $f$ is continuous on $[0,1]$ (a compact interval) it is also uniformly continuous, that means for each $\epsilon>0$ there is a $\delta >0$ such that $|f(x)-f(y)| < \epsilon$ for all $|x-y| < \delta$.
Let $\epsilon >0$. Choose $n$ so big that $|f(x)-f(y)| < \epsilon$ for all $|x-y| < \frac 1n$ (i.e. choose $n$ bigger than $\frac{1}{\delta}$ where $\delta$ is the corresponding $\delta$ from above for our $\epsilon$) . Then $$\sup_{x \in [0,1]} |f_m(x)-f(x)| < \epsilon$$ for all $m \ge n$, because the expression $f_m(x)-f(x)$ is the same as $f(\zeta_k)-f(x)$ for some $\zeta_k$ that satisfies $|\zeta_k - x| < \frac 1m \le \frac 1n$.