Verifying that there is a piecewise linear function on a closed set that approximates it well

Question

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 $$|x-x_k|<\delta.$$

Answer 1

Presumably there should be a hypothesis that $f$ is continuous on $[a,b]$. The argument is almost fine. The one slight hiccup is your assumption that $b-a$ is an integral multiple of $\delta$. After you use continuity of $f$ and compactness of $[a,b]$ to get a $\delta$ that ‘works’ for that $\epsilon$, you want to choose $N$ large enough so that $(b-a)/N<\delta$ and then replace $\delta$ by $(b-a)/N$.