Question: Let $f$ be a continuous function on $[a,b]$. Show that there is a piecewise linear function $\phi$ on $[a,b]$ with $|f(x)-\phi(x)| < \varepsilon$ for $x \in [a,b]$.
My proof: For any $\varepsilon > 0$ there is a $\delta >0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$ for all $x,y \in [a,b]$. Let $b=a+N\delta$ and partition the interval into subintervals as follows: $x_n=a+n\delta$ for $n=1,2,3,...,N$. Now let $f(x_i)=\phi(x_i)$ and let $\phi(x)$ be piecewise linear between $(x_i,x_{x+1})$.
Then for $x \in (x_i, x_{i+1})$: $|f(x)-\phi(x)|=|f(x)-f(x_k)+\phi(x_k)-\phi(x)|\leq|f(x)-f(x_k)|+|\phi(x_k)+\phi(x)| < 2\epsilon,$ since $|x-x_k|<\delta.$