Given $f$ continuous and $\epsilon>0$, prove that exists a polygonal function $g$ such that $|f(x)-g(x)|<\epsilon$ for every $x \in [a,b]$

Question

Definition: A continuous function $\varphi: [a,b]\rightarrow \mathbb{R}$ is called $polygonal$ when exists $a=a_0

Question itself: Let $f:[a,b]\rightarrow \mathbb{R}$ is continuous. Given $\epsilon>0$, prove that there exists a polygonal function $\varphi:[a,b]\rightarrow \mathbb{R}$ such that $|f(x)-g(x)|<\varepsilon$ for every $x\in [a,b]$.

I made some progress, but i think it's not so rigorous:

Let $f:[a,b] \rightarrow \mathbb{R}$ continuous and $\varepsilon>0$ be given. We construct such a polygonal $\varphi$. For partitionate the interval $[a,b]$: $a=a_0<...

We define $\varphi$ inductively. For $[a_0,a_1]$, let $f_0(x_1)$ and $f_0(x_2)$ be the minimum and maximum, respecively. If $|f_0(x_2) - f_0(x_1)|<\varepsilon$, define $\varphi|_{[a_{0},a_1]}$ as the line (polynomial with degree 1) wich joints $f_0(x_1)$ and $f_0(x_2).$Then we do the same process for the interval $[x_2,a_1]$. If $|f_0(x_2) - f_0(x_1)|\geq \varepsilon$, choose any $f_0(x_3)\in f([a_0,a_1])$ such that $|f_0(x_2) - f_0(x_3)|<\varepsilon$, then define for $[x_3,a_1]$. Defined for $[a_2,a_3],...,[a_{n-2},a_{n-1}]$, we define for $[a_{n-1},a_{n}]$ the same way. Hence, $\varphi$ is constructed inductively and continuous, since any polynomial is continuous. Also $|f(x)-\varphi(x)|<\varepsilon$ for every $x$ in the interval $[a,b]$.

Any hints on how to make this more rigorous, is there something that i'm missing?